http://drorbn.net/api.php?action=feedcontributions&user=Katiemann&feedformat=atomDrorbn - User contributions [en]2024-03-29T14:50:02ZUser contributionsMediaWiki 1.21.1http://drorbn.net/index.php?title=0708-1300/Class_notes_for_Tuesday,_November_270708-1300/Class notes for Tuesday, November 272008-01-06T15:38:17Z<p>Katiemann: /* Class Notes */</p>
<hr />
<div>{{0708-1300/Navigation}}<br />
==Today's Agenda==<br />
* The planimeter with a picture from http://whistleralley.com/planimeter/planimeter.htm but our very own plane geometry and Stokes' theorem.<br />
* Completion of the proof of Stokes' theorem.<br />
* Completion of the discussion of the two- and three-dimensional cases of Stokes' theorem.<br />
* With luck, a discussion of de-Rham cohomology, homotopy invariance and Poincaré's lemma.<br />
<br />
<br />
==Class Notes==<br />
<br />
<span style="color: red;">The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.</span><br />
<br />
===First Hour===<br />
<br />
'''Planimeter'''<br />
<br />
A planimeter consists of two rods connected with a join where the end of one rod is fixed (but free to rotate) and the opposing end of the second rod traces out the boundary of some surface on the plane. I.e., the planimeter is kind of like a 1 legged roach. At the join of the two rods is a wheel which rotates (and measures the rotation) when the rod tracing the boundary moves in the normal direction and simply slides back an forth when moved in a tangential direction. <br />
<br />
Now we recall from plane geometry that we can locate points in the polar form <math>(r,\theta)</math> and have the equations <math>x = rcos\theta</math> and <math>y = rsin\theta</math><br />
<br />
Hence, <br />
<br />
<math>dx = cos\theta dr - rsin\theta d\theta</math><br />
<br />
<math>dy = sin\theta dr + rcos\theta d\theta</math><br />
<br />
Hence <math>dx\wedge dy = r(cos^2\theta + sin^2\theta)dr\wedge d\theta = rdr\wedge d\theta</math><br />
<br />
<br />
<br />
<br />
Now, the planimeter is essentially a 1 form corresponding to the speed of the wheel. We consider a diagram where the angle from the horizontal at the fixed end of the planimeter to the measuring end is <math>\theta</math> and the angle from the horizontal to the first rod (the one connected to the fixed point) is <math>\theta + \phi</math>. Hence <math>r = 2cos\phi</math> and <math>dr = -2sin\phi d\phi</math><br />
<br />
With a little plane geometry we can see that <math>\omega = cos2\phi d(\theta + \phi)</math><br />
<br />
Computing, <br />
<br />
<math>d\omega = -2 sin2\phi d\phi\wedge(d\theta + d\phi) = -4 sin\phi cos\phi d\phi\wedge d\theta = 2cos\phi dr\wedge d\theta = rdr\wedge d\theta = dx\wedge dy</math><br />
<br />
<br />
Now applying stokes theorem, the the planimeter integrates <math>\omega</math> over the boundary of our surface and hence this is just the integral of <math>d\omega</math> over the surface. But this is just the integral of the area form. <br />
<br />
Hence the planimeter measure the area of a surface. <br />
<br />
<br />
'''Back to Stokes Theorem'''<br />
<br />
<br />
Firstly recall that <math>\partial M</math> is oriented so that if you prepend the outward normal to its orientation you get the orientation of M<br />
<br />
Alternatively we recall that neighborhoods of points on the boundary look like the half space. Hence we can choose to restrict our attention to atlas's where all charts look like <math>H= \{x\in\mathbb{R}^n:x_1\leq 0\}</math><br />
<br />
We can see that these orientations are the same, i.e., just prepend the outward normal to the half space. <br />
<br />
<br />
''Proof of Stokes''<br />
<br />
<br />
We have now defined all the terms. WLOG <math>\omega</math> is supported in one chart (by linearity)<br />
<br />
For a compactly supported n-1 form on H need to show that <math>\int_{\partial H}\omega = \int_H d\omega</math><br />
<br />
<br />
We let <math>\omega = \sum f_i dx_1\wedge\cdots\wedge \hat{dx_i}\wedge\cdots\wedge dx_n</math> (where the hat means it is omitted)<br />
<br />
<math>d\omega = \sum (-1)^{i-1} \frac{\partial f_i}{\partial x_i} dx_1\wedge\cdots\wedge dx_n</math><br />
<br />
<br />
So, <math>\int_H d\omega = \sum \int_{[x_1\leq 0]}(-1)^{i-1}\frac{\partial f_i}{\partial x_i}dx_1\wedge\cdots\wedge dx_n = \sum (-1)^{i-1}\int_{[x_1\leq 0]}\frac{\partial f_i}{\partial x_i}</math> <br />
<br />
via fundamental theorem of calculus and that the f_i's are compactly supported we get <br />
<br />
<math>= \int_{[x_1\leq 0]} \frac{\partial f_1}{\partial x_1} = \int_{[x_1=0]} f_1</math><br />
<br />
<br />
Hence with the standard inclusion of <math>\partial H = \mathbb{R}^{n-1}_{x_2\cdots x_n}</math> we get <br />
<br />
<br />
<math>\int_{\partial H}\omega = \int_{\mathbb{R}^{n-1}_{x_2\cdots x_n}}\iota^*(\sum f_i dx_1\wedge\cdots\wedge \hat{dx_i}\wedge\cdots\wedge dx_n) = \int_{[x_1=0]}f_1</math><br />
<br />
<br />
Thus these are the same and the theorem is proved ''Q.E.D.''<br />
<br />
<br />
'''Real Plane'''<br />
<br />
Consider <math>\Omega^0(\mathbb{R}^{n-1}_{x_2\cdots x_n})\rightarrow^d\Omega^1(\mathbb{R}^{n-1}_{x_2\cdots x_n})\rightarrow^d\Omega^2(\mathbb{R}^{n-1}_{x_2\cdots x_n})</math><br />
<br />
<br />
Forms in <math>\Omega^1(\mathbb{R}^{n-1}_{x_2\cdots x_n})</math> look like <math>Fdx +Gdy</math> and map under d to <math>(G_x - F_y)dx\wedge dy</math><br />
<br />
Hence applying Stokes' Theorem:<br />
<br />
<math>\int_{\partial D}Fdx + Gdy = \int_D (G_x-F_y)dxdy</math><br />
<br />
This is known as Greens Theorem<br />
<br />
<br />
In complex analysis we also have a similar result Cauchy's Theorem where the integral of an analytic function around a closed path is zero. This is because analytic functions obey the Cauchy-Riemann equations and hence <math>G_x-F_y</math> is identically zero. <br />
<br />
<br />
===Second Hour===<br />
<br />
<br />
'''Example 2'''<br />
<br />
<math>\Omega^k(\mathbb{R}^3)</math><br />
<br />
<br />
Recall previous we had consider the spaces <math>\Omega^k(\mathbb{R}^3)</math> and showed that <math>\Omega^0</math> and <math>\Omega 3</math> corresponded with functions and that <math>\Omega^2</math> and <math>\Omega^1</math> corresponded with triples of functions (i.e. vector fields). We also showed that the d function between these spaces was the gradient, curl and divergence functions from vector calculus. <br />
<br />
We are now interested in integrating, using Stokes Theorem, forms in these spaces. <br />
<br />
<br />
First, note that to a 0 manifold, assigning an orientation to the manifold is just assigning a plus or minus sign to the manifold as a result of it having a trivial basis. <br />
<br />
This is consistent with 0 manifolds being the boundary of 1 manifolds. Indeed, <br />
<br />
<math>\int_{\pm p_0}\omega_0 = \sum\pm f(p_i)</math><br />
<br />
<br />
Now consider a path <math>\gamma:[0,1]\rightarrow\mathbb{R}^3</math><br />
<br />
<math>\int_{\gamma}\omega_1 = \int_{[0,1]}\gamma^*\omega_1 = \int_{[0,1]}\sum f_i d\gamma^*(x_i) = \int_{[0,1]}\sum f_i d\gamma_i</math><br />
<br />
<math>= \int_{[0,1]}\sum f_i\dot{\gamma}_i dt = \int_{\gamma}\vec F\cdot \vec T_{\gamma}</math><br />
<br />
<br />
Now lets compute <math>\omega_2(v,w)</math><br />
<br />
First, <math>dx_2\wedge dx_3 (v,w) = v_2 w_3 - v_3 w_2</math><br />
<br />
Likewise for each component of <math>\omega_2</math> we thus get <br />
<br />
<math>\omega_2(v,w) = \vec G(p)\cdot (v\times w)</math> where <math>\vec G(p)</math> is the vector of coefficients of <math>\omega_2</math><br />
<br />
Now we know that <math>v\times w</math> is a vector perpendicular to v and w with magnitude equal to the area of the defined parallelogram. So, <br />
<br />
<math>\int_{\Sigma}\omega_2 = \int_{\Sigma} \vec G\cdot \vec n d\sigma</math><br />
<br />
where <math>\vec n</math> denotes the normal vector and <math>d\sigma</math> is the area form and <math>\Sigma</math> is a surface<br />
<br />
<br />
Now for <math>\omega_3</math>, <br />
<br />
<br />
<math>\int_D \omega_3 = \int_D g</math><br />
<br />
<br />
now, <math>f(\gamma(1)) - f(\gamma(0)) = \int_{\gamma} (grad\ f)\cdot\vec T</math><br />
<br />
and <math>\int_D div\ G = \int_{\partial D} G\cdot\vec n d\sigma</math> <br />
<br />
<br />
This is Gauss' Divergence Theorem. <br />
<br />
We can think about this as saying that the flow from each point in a domain, when summed up, will be just the flow out of the boundary of the domain. <br />
<br />
<br />
We also get Stokes' Theorem: <br />
<br />
<br />
<math>\int_{\partial\Sigma} F\cdot\vec T = \int_{\Sigma} curl\ F\cdot\vec n d\sigma</math><br />
<br />
<br />
''End of Example''<br />
<br />
<br />
We recall that since <math>d^2 = 0</math>, if <math>\omega = d\lambda</math> then <math>d\omega = 0</math>. But is the converse true? The following Lemma says 'yes', if the domain is <math>\mathbb{R}^n</math><br />
<br />
<br />
'''Poincare's Lemma'''<br />
<br />
On <math>\mathbb{R}^n, d\omega = 0</math> iff <math>\exists\lambda</math> such that <math>\omega = d\lambda</math><br />
<br />
<br />
This is NOT true for general M, as our homework assignment showed since we had a form <math>d\theta</math> that had <math>d(d\theta) = 0</math> but was not d of a form. <br />
<br />
<br />
Likewise, on <math>\mathbb{R}^n-\{0\}</math> we have<br />
<math><br />
\omega = \frac{1}{||x||^{\alpha}}\sum_{i=1}^{n}x_i dx_1\wedge\cdots\wedge\hat{dx_i}\wedge\cdots\wedge dx_n \in\Omega^{n-1}(\mathbb{R}^n-\{0\})</math><br />
<br />
<br />
Claim: <br />
<br />
For appropriate <math>\alpha,\ d\omega = 0</math> but <math>\exists</math> no <math>\lambda</math> such that <math>d\lambda = \omega</math><br />
<br />
This is in our next homework assignment. <br />
<br />
<br />
Now, if there was such a <math>\lambda</math>, <math>\int_{\Sigma}\omega = \int_{\sigma}d\lambda = \int_{\partial\Sigma}\lambda = 0</math><br />
<br />
If <math>\partial\Sigma = \empty</math> (such as any sphere)<br />
<br />
But, <math>\int_{S^2}\omega = 4\pi</math><br />
<br />
<br />
<br />
'''Definition'''<br />
<br />
<math>Z^k(M) := ker d|_{\Omega^k(M)}</math><br />
<br />
<math><br />
B^k(M) := im d|_{\Omega^{k-1}(M)}</math><br />
<br />
<br />
Clearly <math>B^k\subset Z^k</math> so the following definition makes sense:<br />
<br />
<br />
'''Definition''' (de-Rham Cohomology)<br />
<br />
<math>H^k(M):= Z^k(M)/B^k(M)</math><br />
<br />
<br />
'''Claim'''<br />
<br />
<math>H^k(\mathbb{R}^n) = 0</math> yet <math>H^1(S^1)\neq 0</math>. <br />
<br />
Also, <math>H^{n-1}(\mathbb{R}^n - \{x\})\neq 0</math><br />
<br />
<br />
----<br />
<br />
<br />
===More fun with the planimeter===<br />
<br />
<br />
you can build your own low-tech planimeter out of lego!<br />
[[image:0708-1300planimeter1.jpg]]<br />
<br />
<br />
Here my left hand is holding the fixed point steady while the right hand traces the shape. The net number of times the wheel turns while the shape is traced out indicates the area. <br />
<br />
<br />
'''The empirical test:'''<br />
the two shapes below have the same area, and while tracing them the wheel turned almost exactly the same number of times (about 2.3 rotations). Moreover, tracing smaller shapes caused it to turn fewer times, larger shapes more. <br />
[[image:0708-1300planimeter2.jpg]]<br />
<br />
<br />
'''The problem:''' I couldn't do better than this. Calibrating it (calculating how many rotations corresponds to exactly what area) was a nightmare. Mathematically it was possible, but my planimeter is not accurate enough to agree with my math. If you wish to build your own more accurate one, you might want to try using a thinner wheel or one that grips the table better so you don't accidently lose any turning motion.</div>Katiemannhttp://drorbn.net/index.php?title=File:0708-1300planimeter2.jpgFile:0708-1300planimeter2.jpg2008-01-06T15:30:15Z<p>Katiemann: </p>
<hr />
<div></div>Katiemannhttp://drorbn.net/index.php?title=File:0708-1300planimeter1.jpgFile:0708-1300planimeter1.jpg2008-01-06T15:25:58Z<p>Katiemann: </p>
<hr />
<div></div>Katiemannhttp://drorbn.net/index.php?title=0708-1300/Term_Exam_10708-1300/Term Exam 12007-11-19T23:09:10Z<p>Katiemann: /* The Results */</p>
<hr />
<div>{{0708-1300/Navigation}}<br />
<br />
Term Exam 1 took place on Thursday November 8, 2007, at 6PM, at Sydney Smith 1084.<br />
<br />
=={{Dror}}'s Internal Notes==<br />
[[Image:0708-1300-Term Exam I Plan.png|center|480px]]<br />
<br />
==The Exam==<br />
===Front===<br />
<br />
<center><br />
'''Do not turn this page until instructed.'''<br />
<br />
'''Math 1300 Geometry and Topology'''<br />
<br />
'''Term Test'''<br />
<br />
''University of Toronto, November 8, 2007''<br />
<br />
Solve the 4 problems on the other side of this page.<br />
<br />
Each problem is worth 30 points.<br />
<br />
You have two hours to write this test.<br />
<br />
</center><br />
<br />
'''Notes.'''<br />
* No outside material other than stationary is allowed.<br />
* '''Neatness counts! Language counts!''' The ''ideal'' written solution to a problem looks like a page from a textbook; neat and clean and made of complete and grammatical sentences. Definitely phrases like "there exists" or "for every" cannot be skipped. Lectures are mostly made of spoken words, and so the blackboard part of proofs given during lectures often omits or shortens key phrases. The ideal written solution to a problem does not do that.<br />
<br />
<center>'''Good Luck!'''</center><br />
<br />
===Back===<br />
<br />
'''Solve the following 4 problems.''' Each problem is worth 30 points. You have two hours. '''Neatness counts! Language counts!'''<br />
<br />
'''Problem 1 "Compute".''' Let <math>\phi:{\mathbb R}^2_{x,y}\to{\mathbb R}^2_{u,v}</math> be given by <math>u(x,y)=x^2-y^2</math> and <math>v(x,y)=2xy</math>, let <math>f:{\mathbb R}^2_{u,v}\to{\mathbb R}</math> be given by <math>f(u,v)=u^2+v^2</math>, and let <math>\xi\in T_{(0,1)}{\mathbb R}^2_{x,y}</math> be <math>\xi=\partial/\partial x</math>. Compute<br />
the following quantities (with at least some justification):<br />
# <math>\phi_\star\xi</math>.<br />
# <math>\phi^\star f</math>.<br />
# <math>(\phi_\star\xi)f</math>.<br />
# <math>\xi(\phi^\star f)</math>.<br />
<br />
'''Problem 2 "Reproduce".''' The tangent space <math>T_0{\mathbb R}^n</math> to <math>{\mathbb R}^n</math> at <math>0</math> can be defined in the following two ways:<br />
# <math>T_0^1{\mathbb R}^n</math> is the set of all smooth curves <math>\gamma:{\mathbb R}\to{\mathbb R}^n</math> satisfying <math>\gamma(0)=0</math>, modulo the equivalence relation <math>\sim</math>, where <math>\gamma_1\sim\gamma_2</math> iff <math>\dot\gamma_1(0)=\dot\gamma_2(0)</math>, where in general, <math>\dot\gamma</math> denotes the derivative of <math>\gamma(t)</math> with respect to <math>t</math>.<br />
# <math>T_0^2{\mathbb R}^n</math> is the set of all linear functionals <math>D</math> on the vector space of smooth functions on <math>{\mathbb R}^n</math>, which also satisfy Leibnitz' rule, <math>D(fg)=(Df)g(0)+f(0)(Dg)</math>.<br />
<br />
Prove that these two definitions are equivalent (i.e., that there is a natural bijection between <math>T_0^1{\mathbb R}^n</math> and <math>T_0^2{\mathbb R}^n)</math>. If you use a non-trivial lemma from calculus, state it precisely but you don't need to prove it.<br />
<br />
'''Problem 3 "Think".''' Let <math>f:M\to M</math> be a smooth function from a compact manifold <math>M</math> to itself. Prove that there is a point <math>y\in M</math> so that <math>f^{-1}(y)</math> is finite. (In fact, there are many such points).<br />
<br />
'''Problem 4 "Sketch".''' Sketch to the best of your understanding the proof of the Whitney embedding theorem, paying close attention to what is important and little attention to what is not. Here, more than anywhere else, '''neatness and language count!'''<br />
<br />
<center>'''Good Luck!'''</center><br />
<br />
==The Results==<br />
<br />
32 students took this test. The average grade was 82.6 and the standard deviation 24.6. All grades are on [http://ccnet3.utoronto.ca/20079/mat1300yy/ CCNet].<br />
<br />
{{Dror/Students_Divider}}<br />
<br />
'''Some (not necessarily correct) answers to the test:''' <br />
<br />
Problem 1:<br />
[[Image:0708-1300TE1problem1.jpg|100px]]<br />
<br />
Problem 2: <br />
<br />
can be found here: [[0708-1300/Class_notes_for_Thursday%2C_September_27]] <br />
<br />
Problem 3: <br />
[[Image:0708-1300TE1problem3.jpg|100px]]<br />
<br />
Problem 4:<br />
<br />
can be found is in the class notes starting here [[0708-1300/Class_notes_for_Tuesday%2C_October_23]]<br />
<br />
If my solutions to 1 and 3 are incorrect, please delete the files and add your own!<br />
<br />
==Some Additional Reading==<br />
<br />
There are some [http://ocw.mit.edu/OcwWeb/Mathematics/18-965Fall-2004/LectureNotes/index.htm lectures notes] of the [http://ocw.mit.edu/OcwWeb/Mathematics/18-965Fall-2004/CourseHome/index.htm MIT Open Course Ware]. This can be an additional reading for us. There are some [http://ocw.mit.edu/OcwWeb/Mathematics/18-965Fall-2004/Assignments/index.htm exercises] too.<br />
<br />
More [http://www.maths.tcd.ie/~zaitsev/ln.pdf lectures notes] from the University of Dublin. This one has exercises.<br />
<br />
From [http://www.mat.univie.ac.at/~michor/dgbook.pdf Wien] with exercises too.</div>Katiemannhttp://drorbn.net/index.php?title=File:0708-1300TE1problem3.jpgFile:0708-1300TE1problem3.jpg2007-11-19T23:05:51Z<p>Katiemann: </p>
<hr />
<div></div>Katiemannhttp://drorbn.net/index.php?title=File:0708-1300TE1problem1.jpgFile:0708-1300TE1problem1.jpg2007-11-19T23:05:41Z<p>Katiemann: </p>
<hr />
<div></div>Katiemannhttp://drorbn.net/index.php?title=0708-1300/Class_notes_for_Thursday,_September_200708-1300/Class notes for Thursday, September 202007-11-05T15:43:25Z<p>Katiemann: /* a non combinatorial, cut-and-paste topology solution */</p>
<hr />
<div>{{0708-1300/Navigation}}<br />
==Dror's Note==<br />
Come to {{Home Link|Talks/Fields-0709/|my talk}} today at 4:30PM at the Fields Institute!<br />
<br />
{{0708-1300/Class Notes}}<br />
<br />
PDF file of the class notes typed up in latex can be located [http://individual.utoronto.ca/tbazett/mat1300/0708-1300-Class_Notes_latex_20-09.pdf here]<br />
<br />
Tex version of the file is also avaliable [http://individual.utoronto.ca/tbazett/mat1300/0708-1300-Class_Notes_latex_20-09.tex here] so that people can easily make changes and repost here if they wish.<br />
<br />
==Exercise==<br />
{{0708-1300/Solution Warning}}<br />
<br />
Configurations of a Generalized Cockroach (not entirely rigourous) <br />
<br />
Let <math>C_n</math> be the manifold of configurations of a "cockroach" with <math>n</math> legs:<br />
[[Image:0708-1300-cockroach-labelling.jpg||center|200px|]]<br />
<br />
Q: What is <math>C_n</math>?<br />
In particular, what is <math>C_6</math>?<br />
<br />
<math>n=2</math>: Consider <math>C_2</math>. <br />
[[Image:0708-1300-cockroach-C2.jpg||left||200px]]<br />
As in the picture, label the angles of the joints <math>\theta_1</math> and <math>\theta_2</math> and the distances from the body to the feet <math>d_1</math> and <math>d_2</math> respectively. <br />
<br />
The value of <math>\theta_i</math> determines the value of <math>d_i</math>.<br />
So, given values <math>(\theta_1, \theta_2)</math>, possible configurations are given by positions of the body, which must have distance <math>d_i</math> from the <math>i</math>th foot.<br />
<br />
That is, the body must lie on the intersection of the two circles of radius <math>d_i</math> centred at the <math>i</math>th foot:<br />
[[Image:0708-1300-cockroach-twocircles.jpg||center||]]<br />
<br />
<br />
There are <math>0, 1, </math> or <math>2</math> solutions for the body position.<br />
If we look only at the top body position, the pair <math>(\theta_1, \theta_2)</math> determines a unique configuration.<br />
So, we can plot the subset on <math>\R^2_{(\theta_1,\theta_2)}</math>:<br />
[[Image:0708-1300-cockroach-C2region.jpg||center||]]<br />
The boundary points are where the "top" solution is in fact the unique solution.<br />
<br />
By symmetry, taking the bottom solution gives us a similar region, and since the boundaries are where the top and bottom solutions coincide (there is only one solution along the boundary), the entire manifold is given by gluing the boundaries together.<br />
This gives a sphere.<br />
<br />
<math>n=3</math>: Configurations with <math>3</math> legs consist of a configuration with <math>2</math> legs plus the configuration of the third leg.<br />
The configuration of the first <math>2</math> legs fixes the position of the body - and thus, the distance <math>d_3</math> from the third foot to the body.<br />
<br />
For certain configurations of the first two legs, the body is too far from the third foot, so these are not found as part of configurations with three feet.<br />
When the distance from the body to the third foot equals the length of the third leg completely extended, this gives a unique configuration of the <math>3</math>-legged cockroach.<br />
Any closer and there are two possible configurations, corresponding to the two ways that the third joint can bend.<br />
<br />
Let <math>R</math> be the region in <math>C_2 = S_2</math> where the distance to the third leg is close enough to give solutions.<br />
The boundary of <math>R</math> is a curve, consisting of all the points at which there is a unique solution:<br />
[[Image:0708-1300-cockroach-regionR.jpg||center||]]<br />
So the manifold <math>C_3</math> is given by taking two copies of <math>R</math> and gluing their boundaries together.<br />
This gives a sphere.<br />
<br />
<math>n \geq 3</math>: Likewise, given that <math>C_n = S_2</math>, it follows that <math>C_{n+1} = S_2</math> also.<br />
In particular, <math>C_6 = S_2</math>.<br />
<br />
===Dror's Evaluation===<br />
The solution for <math>n=2</math> seems right but hard to understand. For <math>n=3</math> the solution is absolutely right. Unfortunately for <math>n>3</math> the solution is wrong. (Why?)<br />
<br />
--[[User:Drorbn|Drorbn]] 19:46, 24 September 2007 (EDT)<br />
<br />
===A combinatorial solution to the roach problem===<br />
<br />
The answer: <br />
An n-legged roach has configuration space = a torus with <math>1+2^{n-3}(n-4)</math> holes!<br />
<br />
The reason: <br />
The configuration space of a roach can be represented as a graph as follows. Each face is the domain the body moves in when we fix which way each leg is bent. Since each of the n legs can be bent in 2 directions, we have <math>2^n</math> faces. Each edge is the boundary where one leg is straight (it borders the face where the leg is bent in the other direction). The vertices therefore correspond to positions where two adjacent legs are straight. This means that each vertex has degree four. (for those of you unfamiliar with graph theory lingo, degree 4 means that we have 4 edges from each vertex. In this case we have also four faces) We can now calculate the number of edges, vertices and faces (E, V, F) of an n-legged roach: <br />
<br />
F = <math>2^n</math><br />
<br />
E = <math>nF/2</math> = <math>2^{n-1}n</math><br />
<br />
V = <math>2E/degree</math> = <math>2^{n-2}n</math><br />
<br />
(the above calculation of E in terms of F and V in terms of E is a fun counting argument, if you can’t figure it out ask me)<br />
Now we compute the Euler genus which tells us what kind of surface this graph is:<br />
Euler genus = <math>2 - V + E - F = 2 + 2^{n-2}(n-4)</math><br />
Since a k-holed torus has genus 2k (the sphere has genus 0), the configuration space of an n-legged roach is a <br />
k = <math>1 + 2^{n-3}(n-4)</math> -holed torus.<br />
<br />
===a non combinatorial, cut-and-paste topology solution===<br />
which, conveniently, gives the same answer as my combinatorial solution above.<br />
[[Image:0708-1300roach1.jpg|thumb|center|200px]]<br />
[[Image:0708-1300roach2001.jpg|thumb|center|200px]]<br />
please e-mail me if you have questions, comments or a burning desire for more rigour.<br />
<br />
a webpage to go to if you want info about how to turn my recursive formula for Cn into my non-recursive one is:<br />
http://web.cs.wpi.edu/~cs504/s99/notes/recurrence/solve/solve.html<br />
<br />
--[[User:Katiemann|Katiemann]] 10:42, 5 November 2007 (EST)<br />
<br />
==Enrichment: Configuration Spaces of Mechanical Systems==<br />
As Prof. Bar-Natan mentioned in class, manifolds are quite useful for studying the physics of mechanical systems like robots and steam engines. There is actually a very general mathematical theory of rigid body mechanics that makes plenty of use of differential geometry. See, for example, <br />
{{ref|BulloLewis_04}} for a discussion of one of these models and its application to nonlinear control theory for mechanical systems. The following is a brief introduction to the idea of a mechanical system's <i>configuration space</i>, and makes use of the notation of {{ref|BulloLewis_04}}.<br />
<br />
Imagine that you have some rigid body <math>B\!</math> floating around in space. The first thing to do is to set up an orthonormal frame <math>(O_{spatial},(s_1,s_2,s_3))\!</math> with respect to which we will describe <math>B\!</math>'s position. Here <math>O_{spatial}\!</math> is some point in Euclidean (affine) space which serves as the origin of our coordinate system and <math>(s_1,s_2,s_3)\!</math> is an orthonormal basis for Euclidean space centred at <math>O_{spatial}\!</math>.<br />
<br />
Now suppose that we rigidly attach a frame <math>(O_{body},(b_1,b_2,b_3))\!</math> to <math>B\!</math> in the sense that the frame's origin is fixed to a specific point on the body (such as the centre of mass) and the basis <math>(b_1,b_2,b_3)\!</math> rotates around with <math>B\!</math>. We can now specify <math>B\!</math>'s configuration with the vector<br />
<br />
::<math>r = O_{body} - O_{spatial} \in \mathbb{R}^3\!</math><br />
<br />
and the matrix <math> R \in SO(3;\mathbb{R})\!</math> which gives the orientation of the basis <math>(b_1,b_2,b_3)\!</math> relative to <math>(s_1,s_2,s_3)\!</math>. Here <math>SO(3;\mathbb{R})\!</math> is the group of <math>3\times 3\!</math> real matrices with unit determinant---the group of rotations of <math>\mathbb{R}^3\!</math>. Concisely, the configuration space of a rigid body is the space <math>Q_{free}=SO(3;\mathbb{R}) \times \mathbb{R}^3\!</math>.<br />
<br />
Lucky for us, <math>SO(3;\mathbb{R})\!</math> is actually a manifold. It's an example of something called a <i>Lie group</i>, which is a group that is also a manifold and whose group operations are smooth according to the differentiable structure of the manifold. Hence <math>Q_{free}\!</math> is also a manifold.<br />
<br />
Now imagine we have <math>k\!</math> rigid bodies. If the bodies are free to move around independently, and if we briefly suspend disbelief and allow them to pass right through each other instead of colliding, our configuration space is <math>Q_{free}^k= \left(SO(3;\mathbb{R}) \times \mathbb{R}^3\right)^k</math>. If, on the other hand, the bodies are interconnected---ie., their motions are not independent---or if the motion of the bodies is constrained in some other way, then the configuration space for the system will be some subset <math>Q \subset Q_{free}^k\!</math>. In fact, it is (usually) a <i>submanifold</i>. <br />
<br />
This situation is precisely what's happening with the roach's configuration. In this case the system consists of 12 rigid bodies (six legs composed of two segments each) that are constrained to move in a plane, connected together at their ends, and pinned down.<br />
<br />
Hopefully, it is starting to become clear how we can model some of the physics now: we can describe the motion of a mechanical system with a curve <math>\gamma : I \rightarrow Q\!</math> where <math>I \subset \mathbb{R}\!</math> is an interval. The velocities of all the bodies (both translational and rotational) at time <math>t\!</math>are encapsulated in the tangent vector <math>\gamma'(t) \in T_{\gamma(t)}Q\!</math>. Similarly, there are nice differential-geometric ways of describing the kinetic and potential energies, forces and so on.<br />
<br />
==References==<br />
{{note|BulloLewis_04}}Francesco Bullo and Andrew D. Lewis, <i>Geometric Control of Mechanical Systems: Modeling, Analysis, and Design for Simple Mechanical Control Systems</i>, Springer-Verlag, New York-Heidelberg-Berlin, 2004. Number 49 in <i>Texts in Applied Mathematics.</i></div>Katiemannhttp://drorbn.net/index.php?title=0708-1300/Class_notes_for_Thursday,_September_200708-1300/Class notes for Thursday, September 202007-11-05T15:42:05Z<p>Katiemann: /* a non combinatorial, cut-and-paste solution */</p>
<hr />
<div>{{0708-1300/Navigation}}<br />
==Dror's Note==<br />
Come to {{Home Link|Talks/Fields-0709/|my talk}} today at 4:30PM at the Fields Institute!<br />
<br />
{{0708-1300/Class Notes}}<br />
<br />
PDF file of the class notes typed up in latex can be located [http://individual.utoronto.ca/tbazett/mat1300/0708-1300-Class_Notes_latex_20-09.pdf here]<br />
<br />
Tex version of the file is also avaliable [http://individual.utoronto.ca/tbazett/mat1300/0708-1300-Class_Notes_latex_20-09.tex here] so that people can easily make changes and repost here if they wish.<br />
<br />
==Exercise==<br />
{{0708-1300/Solution Warning}}<br />
<br />
Configurations of a Generalized Cockroach (not entirely rigourous) <br />
<br />
Let <math>C_n</math> be the manifold of configurations of a "cockroach" with <math>n</math> legs:<br />
[[Image:0708-1300-cockroach-labelling.jpg||center|200px|]]<br />
<br />
Q: What is <math>C_n</math>?<br />
In particular, what is <math>C_6</math>?<br />
<br />
<math>n=2</math>: Consider <math>C_2</math>. <br />
[[Image:0708-1300-cockroach-C2.jpg||left||200px]]<br />
As in the picture, label the angles of the joints <math>\theta_1</math> and <math>\theta_2</math> and the distances from the body to the feet <math>d_1</math> and <math>d_2</math> respectively. <br />
<br />
The value of <math>\theta_i</math> determines the value of <math>d_i</math>.<br />
So, given values <math>(\theta_1, \theta_2)</math>, possible configurations are given by positions of the body, which must have distance <math>d_i</math> from the <math>i</math>th foot.<br />
<br />
That is, the body must lie on the intersection of the two circles of radius <math>d_i</math> centred at the <math>i</math>th foot:<br />
[[Image:0708-1300-cockroach-twocircles.jpg||center||]]<br />
<br />
<br />
There are <math>0, 1, </math> or <math>2</math> solutions for the body position.<br />
If we look only at the top body position, the pair <math>(\theta_1, \theta_2)</math> determines a unique configuration.<br />
So, we can plot the subset on <math>\R^2_{(\theta_1,\theta_2)}</math>:<br />
[[Image:0708-1300-cockroach-C2region.jpg||center||]]<br />
The boundary points are where the "top" solution is in fact the unique solution.<br />
<br />
By symmetry, taking the bottom solution gives us a similar region, and since the boundaries are where the top and bottom solutions coincide (there is only one solution along the boundary), the entire manifold is given by gluing the boundaries together.<br />
This gives a sphere.<br />
<br />
<math>n=3</math>: Configurations with <math>3</math> legs consist of a configuration with <math>2</math> legs plus the configuration of the third leg.<br />
The configuration of the first <math>2</math> legs fixes the position of the body - and thus, the distance <math>d_3</math> from the third foot to the body.<br />
<br />
For certain configurations of the first two legs, the body is too far from the third foot, so these are not found as part of configurations with three feet.<br />
When the distance from the body to the third foot equals the length of the third leg completely extended, this gives a unique configuration of the <math>3</math>-legged cockroach.<br />
Any closer and there are two possible configurations, corresponding to the two ways that the third joint can bend.<br />
<br />
Let <math>R</math> be the region in <math>C_2 = S_2</math> where the distance to the third leg is close enough to give solutions.<br />
The boundary of <math>R</math> is a curve, consisting of all the points at which there is a unique solution:<br />
[[Image:0708-1300-cockroach-regionR.jpg||center||]]<br />
So the manifold <math>C_3</math> is given by taking two copies of <math>R</math> and gluing their boundaries together.<br />
This gives a sphere.<br />
<br />
<math>n \geq 3</math>: Likewise, given that <math>C_n = S_2</math>, it follows that <math>C_{n+1} = S_2</math> also.<br />
In particular, <math>C_6 = S_2</math>.<br />
<br />
===Dror's Evaluation===<br />
The solution for <math>n=2</math> seems right but hard to understand. For <math>n=3</math> the solution is absolutely right. Unfortunately for <math>n>3</math> the solution is wrong. (Why?)<br />
<br />
--[[User:Drorbn|Drorbn]] 19:46, 24 September 2007 (EDT)<br />
<br />
===A combinatorial solution to the roach problem===<br />
<br />
The answer: <br />
An n-legged roach has configuration space = a torus with <math>1+2^{n-3}(n-4)</math> holes!<br />
<br />
The reason: <br />
The configuration space of a roach can be represented as a graph as follows. Each face is the domain the body moves in when we fix which way each leg is bent. Since each of the n legs can be bent in 2 directions, we have <math>2^n</math> faces. Each edge is the boundary where one leg is straight (it borders the face where the leg is bent in the other direction). The vertices therefore correspond to positions where two adjacent legs are straight. This means that each vertex has degree four. (for those of you unfamiliar with graph theory lingo, degree 4 means that we have 4 edges from each vertex. In this case we have also four faces) We can now calculate the number of edges, vertices and faces (E, V, F) of an n-legged roach: <br />
<br />
F = <math>2^n</math><br />
<br />
E = <math>nF/2</math> = <math>2^{n-1}n</math><br />
<br />
V = <math>2E/degree</math> = <math>2^{n-2}n</math><br />
<br />
(the above calculation of E in terms of F and V in terms of E is a fun counting argument, if you can’t figure it out ask me)<br />
Now we compute the Euler genus which tells us what kind of surface this graph is:<br />
Euler genus = <math>2 - V + E - F = 2 + 2^{n-2}(n-4)</math><br />
Since a k-holed torus has genus 2k (the sphere has genus 0), the configuration space of an n-legged roach is a <br />
k = <math>1 + 2^{n-3}(n-4)</math> -holed torus.<br />
<br />
===a non combinatorial, cut-and-paste topology solution===<br />
which, conveniently, gives the same answer as the solution above.<br />
[[Image:0708-1300roach1.jpg|thumb|center|200px]]<br />
[[Image:0708-1300roach2001.jpg|thumb|center|200px]]<br />
please e-mail me if you have questions, comments or a burning desire for more rigour.<br />
<br />
a webpage to go to if you want info about how to turn my recursive formula for Cn into my non-recursive one is:<br />
http://web.cs.wpi.edu/~cs504/s99/notes/recurrence/solve/solve.html<br />
<br />
--[[User:Katiemann|Katiemann]] 10:42, 5 November 2007 (EST)<br />
<br />
==Enrichment: Configuration Spaces of Mechanical Systems==<br />
As Prof. Bar-Natan mentioned in class, manifolds are quite useful for studying the physics of mechanical systems like robots and steam engines. There is actually a very general mathematical theory of rigid body mechanics that makes plenty of use of differential geometry. See, for example, <br />
{{ref|BulloLewis_04}} for a discussion of one of these models and its application to nonlinear control theory for mechanical systems. The following is a brief introduction to the idea of a mechanical system's <i>configuration space</i>, and makes use of the notation of {{ref|BulloLewis_04}}.<br />
<br />
Imagine that you have some rigid body <math>B\!</math> floating around in space. The first thing to do is to set up an orthonormal frame <math>(O_{spatial},(s_1,s_2,s_3))\!</math> with respect to which we will describe <math>B\!</math>'s position. Here <math>O_{spatial}\!</math> is some point in Euclidean (affine) space which serves as the origin of our coordinate system and <math>(s_1,s_2,s_3)\!</math> is an orthonormal basis for Euclidean space centred at <math>O_{spatial}\!</math>.<br />
<br />
Now suppose that we rigidly attach a frame <math>(O_{body},(b_1,b_2,b_3))\!</math> to <math>B\!</math> in the sense that the frame's origin is fixed to a specific point on the body (such as the centre of mass) and the basis <math>(b_1,b_2,b_3)\!</math> rotates around with <math>B\!</math>. We can now specify <math>B\!</math>'s configuration with the vector<br />
<br />
::<math>r = O_{body} - O_{spatial} \in \mathbb{R}^3\!</math><br />
<br />
and the matrix <math> R \in SO(3;\mathbb{R})\!</math> which gives the orientation of the basis <math>(b_1,b_2,b_3)\!</math> relative to <math>(s_1,s_2,s_3)\!</math>. Here <math>SO(3;\mathbb{R})\!</math> is the group of <math>3\times 3\!</math> real matrices with unit determinant---the group of rotations of <math>\mathbb{R}^3\!</math>. Concisely, the configuration space of a rigid body is the space <math>Q_{free}=SO(3;\mathbb{R}) \times \mathbb{R}^3\!</math>.<br />
<br />
Lucky for us, <math>SO(3;\mathbb{R})\!</math> is actually a manifold. It's an example of something called a <i>Lie group</i>, which is a group that is also a manifold and whose group operations are smooth according to the differentiable structure of the manifold. Hence <math>Q_{free}\!</math> is also a manifold.<br />
<br />
Now imagine we have <math>k\!</math> rigid bodies. If the bodies are free to move around independently, and if we briefly suspend disbelief and allow them to pass right through each other instead of colliding, our configuration space is <math>Q_{free}^k= \left(SO(3;\mathbb{R}) \times \mathbb{R}^3\right)^k</math>. If, on the other hand, the bodies are interconnected---ie., their motions are not independent---or if the motion of the bodies is constrained in some other way, then the configuration space for the system will be some subset <math>Q \subset Q_{free}^k\!</math>. In fact, it is (usually) a <i>submanifold</i>. <br />
<br />
This situation is precisely what's happening with the roach's configuration. In this case the system consists of 12 rigid bodies (six legs composed of two segments each) that are constrained to move in a plane, connected together at their ends, and pinned down.<br />
<br />
Hopefully, it is starting to become clear how we can model some of the physics now: we can describe the motion of a mechanical system with a curve <math>\gamma : I \rightarrow Q\!</math> where <math>I \subset \mathbb{R}\!</math> is an interval. The velocities of all the bodies (both translational and rotational) at time <math>t\!</math>are encapsulated in the tangent vector <math>\gamma'(t) \in T_{\gamma(t)}Q\!</math>. Similarly, there are nice differential-geometric ways of describing the kinetic and potential energies, forces and so on.<br />
<br />
==References==<br />
{{note|BulloLewis_04}}Francesco Bullo and Andrew D. Lewis, <i>Geometric Control of Mechanical Systems: Modeling, Analysis, and Design for Simple Mechanical Control Systems</i>, Springer-Verlag, New York-Heidelberg-Berlin, 2004. Number 49 in <i>Texts in Applied Mathematics.</i></div>Katiemannhttp://drorbn.net/index.php?title=File:0708-1300roach2001.jpgFile:0708-1300roach2001.jpg2007-11-05T15:36:17Z<p>Katiemann: </p>
<hr />
<div></div>Katiemannhttp://drorbn.net/index.php?title=File:0708-1300roach1.jpgFile:0708-1300roach1.jpg2007-11-05T15:25:12Z<p>Katiemann: </p>
<hr />
<div></div>Katiemannhttp://drorbn.net/index.php?title=0708-1300/Class_notes_for_Thursday,_September_200708-1300/Class notes for Thursday, September 202007-11-02T13:39:10Z<p>Katiemann: /* Exercise */</p>
<hr />
<div>{{0708-1300/Navigation}}<br />
==Dror's Note==<br />
Come to {{Home Link|Talks/Fields-0709/|my talk}} today at 4:30PM at the Fields Institute!<br />
<br />
{{0708-1300/Class Notes}}<br />
<br />
PDF file of the class notes typed up in latex can be located [http://individual.utoronto.ca/tbazett/mat1300/0708-1300-Class_Notes_latex_20-09.pdf here]<br />
<br />
Tex version of the file is also avaliable [http://individual.utoronto.ca/tbazett/mat1300/0708-1300-Class_Notes_latex_20-09.tex here] so that people can easily make changes and repost here if they wish.<br />
<br />
==Exercise==<br />
{{0708-1300/Solution Warning}}<br />
<br />
Configurations of a Generalized Cockroach (not entirely rigourous) <br />
<br />
Let <math>C_n</math> be the manifold of configurations of a "cockroach" with <math>n</math> legs:<br />
[[Image:0708-1300-cockroach-labelling.jpg||center|200px|]]<br />
<br />
Q: What is <math>C_n</math>?<br />
In particular, what is <math>C_6</math>?<br />
<br />
<math>n=2</math>: Consider <math>C_2</math>. <br />
[[Image:0708-1300-cockroach-C2.jpg||left||200px]]<br />
As in the picture, label the angles of the joints <math>\theta_1</math> and <math>\theta_2</math> and the distances from the body to the feet <math>d_1</math> and <math>d_2</math> respectively. <br />
<br />
The value of <math>\theta_i</math> determines the value of <math>d_i</math>.<br />
So, given values <math>(\theta_1, \theta_2)</math>, possible configurations are given by positions of the body, which must have distance <math>d_i</math> from the <math>i</math>th foot.<br />
<br />
That is, the body must lie on the intersection of the two circles of radius <math>d_i</math> centred at the <math>i</math>th foot:<br />
[[Image:0708-1300-cockroach-twocircles.jpg||center||]]<br />
<br />
<br />
There are <math>0, 1, </math> or <math>2</math> solutions for the body position.<br />
If we look only at the top body position, the pair <math>(\theta_1, \theta_2)</math> determines a unique configuration.<br />
So, we can plot the subset on <math>\R^2_{(\theta_1,\theta_2)}</math>:<br />
[[Image:0708-1300-cockroach-C2region.jpg||center||]]<br />
The boundary points are where the "top" solution is in fact the unique solution.<br />
<br />
By symmetry, taking the bottom solution gives us a similar region, and since the boundaries are where the top and bottom solutions coincide (there is only one solution along the boundary), the entire manifold is given by gluing the boundaries together.<br />
This gives a sphere.<br />
<br />
<math>n=3</math>: Configurations with <math>3</math> legs consist of a configuration with <math>2</math> legs plus the configuration of the third leg.<br />
The configuration of the first <math>2</math> legs fixes the position of the body - and thus, the distance <math>d_3</math> from the third foot to the body.<br />
<br />
For certain configurations of the first two legs, the body is too far from the third foot, so these are not found as part of configurations with three feet.<br />
When the distance from the body to the third foot equals the length of the third leg completely extended, this gives a unique configuration of the <math>3</math>-legged cockroach.<br />
Any closer and there are two possible configurations, corresponding to the two ways that the third joint can bend.<br />
<br />
Let <math>R</math> be the region in <math>C_2 = S_2</math> where the distance to the third leg is close enough to give solutions.<br />
The boundary of <math>R</math> is a curve, consisting of all the points at which there is a unique solution:<br />
[[Image:0708-1300-cockroach-regionR.jpg||center||]]<br />
So the manifold <math>C_3</math> is given by taking two copies of <math>R</math> and gluing their boundaries together.<br />
This gives a sphere.<br />
<br />
<math>n \geq 3</math>: Likewise, given that <math>C_n = S_2</math>, it follows that <math>C_{n+1} = S_2</math> also.<br />
In particular, <math>C_6 = S_2</math>.<br />
<br />
===Dror's Evaluation===<br />
The solution for <math>n=2</math> seems right but hard to understand. For <math>n=3</math> the solution is absolutely right. Unfortunately for <math>n>3</math> the solution is wrong. (Why?)<br />
<br />
--[[User:Drorbn|Drorbn]] 19:46, 24 September 2007 (EDT)<br />
<br />
===A combinatorial solution to the roach problem===<br />
<br />
The answer: <br />
An n-legged roach has configuration space = a torus with <math>1+2^{n-3}(n-4)</math> holes!<br />
<br />
The reason: <br />
The configuration space of a roach can be represented as a graph as follows. Each face is the domain the body moves in when we fix which way each leg is bent. Since each of the n legs can be bent in 2 directions, we have <math>2^n</math> faces. Each edge is the boundary where one leg is straight (it borders the face where the leg is bent in the other direction). The vertices therefore correspond to positions where two adjacent legs are straight. This means that each vertex has degree four. (for those of you unfamiliar with graph theory lingo, degree 4 means that we have 4 edges from each vertex. In this case we have also four faces) We can now calculate the number of edges, vertices and faces (E, V, F) of an n-legged roach: <br />
<br />
F = <math>2^n</math><br />
<br />
E = <math>nF/2</math> = <math>2^{n-1}n</math><br />
<br />
V = <math>2E/degree</math> = <math>2^{n-2}n</math><br />
<br />
(the above calculation of E in terms of F and V in terms of E is a fun counting argument, if you can’t figure it out ask me)<br />
Now we compute the Euler genus which tells us what kind of surface this graph is:<br />
Euler genus = <math>2 - V + E - F = 2 + 2^{n-2}(n-4)</math><br />
Since a k-holed torus has genus 2k (the sphere has genus 0), the configuration space of an n-legged roach is a <br />
k = <math>1 + 2^{n-3}(n-4)</math> -holed torus.<br />
<br />
===a non combinatorial, cut-and-paste solution===<br />
will be posted as soon as I can get access to a scanner. In the mean time, I reccommend Diestal's ''Graph Theory'' if you want to know what Euler genus is. <br />
--[[User:Katiemann|Katiemann]] 09:39, 2 November 2007 (EDT)<br />
<br />
==Enrichment: Configuration Spaces of Mechanical Systems==<br />
As Prof. Bar-Natan mentioned in class, manifolds are quite useful for studying the physics of mechanical systems like robots and steam engines. There is actually a very general mathematical theory of rigid body mechanics that makes plenty of use of differential geometry. See, for example, <br />
{{ref|BulloLewis_04}} for a discussion of one of these models and its application to nonlinear control theory for mechanical systems. The following is a brief introduction to the idea of a mechanical system's <i>configuration space</i>, and makes use of the notation of {{ref|BulloLewis_04}}.<br />
<br />
Imagine that you have some rigid body <math>B\!</math> floating around in space. The first thing to do is to set up an orthonormal frame <math>(O_{spatial},(s_1,s_2,s_3))\!</math> with respect to which we will describe <math>B\!</math>'s position. Here <math>O_{spatial}\!</math> is some point in Euclidean (affine) space which serves as the origin of our coordinate system and <math>(s_1,s_2,s_3)\!</math> is an orthonormal basis for Euclidean space centred at <math>O_{spatial}\!</math>.<br />
<br />
Now suppose that we rigidly attach a frame <math>(O_{body},(b_1,b_2,b_3))\!</math> to <math>B\!</math> in the sense that the frame's origin is fixed to a specific point on the body (such as the centre of mass) and the basis <math>(b_1,b_2,b_3)\!</math> rotates around with <math>B\!</math>. We can now specify <math>B\!</math>'s configuration with the vector<br />
<br />
::<math>r = O_{body} - O_{spatial} \in \mathbb{R}^3\!</math><br />
<br />
and the matrix <math> R \in SO(3;\mathbb{R})\!</math> which gives the orientation of the basis <math>(b_1,b_2,b_3)\!</math> relative to <math>(s_1,s_2,s_3)\!</math>. Here <math>SO(3;\mathbb{R})\!</math> is the group of <math>3\times 3\!</math> real matrices with unit determinant---the group of rotations of <math>\mathbb{R}^3\!</math>. Concisely, the configuration space of a rigid body is the space <math>Q_{free}=SO(3;\mathbb{R}) \times \mathbb{R}^3\!</math>.<br />
<br />
Lucky for us, <math>SO(3;\mathbb{R})\!</math> is actually a manifold. It's an example of something called a <i>Lie group</i>, which is a group that is also a manifold and whose group operations are smooth according to the differentiable structure of the manifold. Hence <math>Q_{free}\!</math> is also a manifold.<br />
<br />
Now imagine we have <math>k\!</math> rigid bodies. If the bodies are free to move around independently, and if we briefly suspend disbelief and allow them to pass right through each other instead of colliding, our configuration space is <math>Q_{free}^k= \left(SO(3;\mathbb{R}) \times \mathbb{R}^3\right)^k</math>. If, on the other hand, the bodies are interconnected---ie., their motions are not independent---or if the motion of the bodies is constrained in some other way, then the configuration space for the system will be some subset <math>Q \subset Q_{free}^k\!</math>. In fact, it is (usually) a <i>submanifold</i>. <br />
<br />
This situation is precisely what's happening with the roach's configuration. In this case the system consists of 12 rigid bodies (six legs composed of two segments each) that are constrained to move in a plane, connected together at their ends, and pinned down.<br />
<br />
Hopefully, it is starting to become clear how we can model some of the physics now: we can describe the motion of a mechanical system with a curve <math>\gamma : I \rightarrow Q\!</math> where <math>I \subset \mathbb{R}\!</math> is an interval. The velocities of all the bodies (both translational and rotational) at time <math>t\!</math>are encapsulated in the tangent vector <math>\gamma'(t) \in T_{\gamma(t)}Q\!</math>. Similarly, there are nice differential-geometric ways of describing the kinetic and potential energies, forces and so on.<br />
<br />
==References==<br />
{{note|BulloLewis_04}}Francesco Bullo and Andrew D. Lewis, <i>Geometric Control of Mechanical Systems: Modeling, Analysis, and Design for Simple Mechanical Control Systems</i>, Springer-Verlag, New York-Heidelberg-Berlin, 2004. Number 49 in <i>Texts in Applied Mathematics.</i></div>Katiemannhttp://drorbn.net/index.php?title=0708-1300/Class_Photo0708-1300/Class Photo2007-09-30T15:14:44Z<p>Katiemann: /* Who We Are... */</p>
<hr />
<div>{{0708-1300/Navigation}}<br />
<br />
Our class on September 27, 2007:<br />
<br />
[[Image:0708-1300-ClassPhoto.jpg|thumb|centre|600px|Class Photo: click to enlarge]]<br />
<br />
Please identify yourself in this photo! There are two ways to do that:<br />
<br />
* [[Special:Userlogin|Log in]] to this Wiki and edit this page. Put your name, userid, email address and location in the picture in the alphabetical list below.<br />
* Send [[User:Drorbn|Dror]] an email message with this information.<br />
<br />
The first option is more fun but less private.<br />
<br />
===Who We Are...===<br />
<br />
{| align=center border=1 <br />
|-<br />
!First name<br />
!Last name <br />
!UserID <br />
!Email <br />
!In the photo <br />
!Comments<br />
{{Photo Entry|last=Bar-Natan|first=Dror|userid=Drorbn|email=drorbn @ math.toronto.edu|location=facing everybody, as the photographer|comments=Take this entry as a model and leave it first. Otherwise alphabetize by last name. Feel free to leave some fields blank}}<br />
{{Photo Entry|last=Bazett|first=Trefor|userid=Trefor|email=trefor.bazett @ toronto.ca|location=tallest person a little right of center in a beige shirt|comments=}}<br />
{{Photo Entry|last=Bjorndahl|first=Adam|userid=ABjorndahl|email=adam.bjorndahl @ utoronto.ca|location=back row, fifth from the left, under the "f(tp)dt"|comments=Looking forward to a great year!}}<br />
{{Photo Entry|last=Chow|first=Aaron|userid=aaron.chow|email=aaron @ utoronto.ca|location=Third from right, in a black shirt.|comments=Hope we have a good year together!}}<br />
{{Photo Entry|last=Isgur|first=Abraham|userid=Abisgu|email=abraham.isgur@ math.toronto.edu|location=2nd person in the back row, from the right, the one with the beard and long hair|comments=}}<br />
{{Photo Entry|last=Mann|first=Katie|userid=katiemann|email=katie.mann@ utoronto.ca|location=middle, wearing "Eulers" shirt|comments=}}<br />
{{Photo Entry|last=Pym|first=Brent|userid=Bpym|email=bpym @ math.toronto.edu|location=10th from the right (cumulatively), under the <math>T_p(M)\!</math>|comments=Adding this entry was my first-ever edit of a Wiki!}}<br />
{{Photo Entry|last=Vera Pacheco|first=Franklin|userid=Franklin|email=franklin.vp @ gmail.com|location=Xth from left to right|comments=To find me you must first go to [[http://www.deathball.net/notpron/]] solve the first 4 pages. Once this done you will know how to find me. Once this done go back to NOTPRON an solve the rest of the puzzle}}<br />
{{Photo Entry|last=Watts|first=Jordan|userid=Jwatts|email=jwatts @ math.toronto.edu|location=in the back, 2nd or 3rd from the left, depending on your convention|comments=My glasses become invisible in pictures.}}<br />
{{Photo Entry|last=Wong|first=Silian|userid=kuramay|email=kurama_y @ hotmail.com|location=One of the Asian-looking girls...with sparkling teeth(??)|comments=I'll write up some comments after their existences}}<br />
|}</div>Katiemannhttp://drorbn.net/index.php?title=0708-1300/Class_notes_for_Thursday,_September_200708-1300/Class notes for Thursday, September 202007-09-25T18:00:35Z<p>Katiemann: /* Exercise */</p>
<hr />
<div>{{0708-1300/Navigation}}<br />
==Dror's Note==<br />
Come to {{Home Link|Talks/Fields-0709/|my talk}} today at 4:30PM at the Fields Institute!<br />
<br />
{{0708-1300/Class Notes}}<br />
<br />
PDF file of the class notes typed up in latex can be located [http://individual.utoronto.ca/tbazett/mat1300/0708-1300-Class_Notes_latex_20-09.pdf here]<br />
<br />
Tex version of the file is also avaliable [http://individual.utoronto.ca/tbazett/mat1300/0708-1300-Class_Notes_latex_20-09.tex here] so that people can easily make changes and repost here if they wish.<br />
<br />
==Exercise==<br />
{{07008-1300/Solution Warning}}<br />
<br />
Configurations of a Generalized Cockroach (not entirely rigourous) <br />
<br />
Let <math>C_n</math> be the manifold of configurations of a "cockroach" with <math>n</math> legs:<br />
[[Image:0708-1300-cockroach-labelling.jpg||center|200px|]]<br />
<br />
Q: What is <math>C_n</math>?<br />
In particular, what is <math>C_6</math>?<br />
<br />
p<math>n=2</math>: Consider <math>C_2</math>. <br />
[[Image:0708-1300-cockroach-C2.jpg||left||200px]]<br />
As in the picture, label the angles of the joints <math>\theta_1</math> and <math>\theta_2</math> and the distances from the body to the feet <math>d_1</math> and <math>d_2</math> respectively. <br />
<br />
The value of <math>\theta_i</math> determines the value of <math>d_i</math>.<br />
So, given values <math>(\theta_1, \theta_2)</math>, possible configurations are given by positions of the body, which must have distance <math>d_i</math> from the <math>i</math>th foot.<br />
<br />
That is, the body must lie on the intersection of the two circles of radius <math>d_i</math> centred at the <math>i</math>th foot:<br />
[[Image:0708-1300-cockroach-twocircles.jg||center||]]<br />
<br />
<br />
There are <math>0, 1, </math> or <math>2</math> solutions for the body position.<br />
If we look only at the top body position, the pair <math>(\theta_1, \theta_2)</math> determines a unique configuration.<br />
So, we can plot the subset on <math>\R^2_{(\theta_1,\theta_2)}</math>:<br />
[[Image:0708-1300-cockroach-C2region.jpg||center||]]<br />
The boundary points are where the "top" solution is in fact the unique solution.<br />
<br />
By symmetry, taking the bottom solution gives us a similar region, and since the boundaries are where the top and bottom solutions coincide (there is only one solution along the boundary), the entire manifold is given by gluing the boundaries together.<br />
This gives a sphere.<br />
<br />
<math>n=3</math>: Configurations with <math>3</math> legs consist of a configuration with <math>2</math> legs plus the configuration of the third leg.<br />
The configuration of the first <math>2</math> legs fixes the position of the body - and thus, the distance <math>d_3</math> from the third foot to the body.<br />
<br />
For certain configurations of the first two legs, the body is too far from the third foot, so these are not found as part of configurations with three feet.<br />
When the distance from the body to the third foot equals the length of the third leg completely extended, this gives a unique configuration of the <math>3</math>-legged cockroach.<br />
Any closer and there are two possible configurations, corresponding to the two ways that the third joint can bend.<br />
<br />
Let <math>R</math> be the region in <math>C_2 = S_2</math> where the distance to the third leg is close enough to give solutions.<br />
The boundary of <math>R</math> is a curve, consisting of all the points at which there is a unique solution:<br />
[[Image:0708-1300-cockroach-regionR.jpg||center||]]<br />
So the manifold <math>C_3</math> is given by taking two copies of <math>R</math> and gluing their boundaries together.<br />
This gives a sphere.<br />
<br />
<math>n \geq 3</math>: Likewise, given that <math>C_n = S_2</math>, it follows that <math>C_{n+1} = S_2</math> also.<br />
In particular, <math>C_6 = S_2</math>.<br />
<br />
===Dror's Evaluation===<br />
The solution for <math>n=2</math> seems right but hard to understand. For <math>n=3</math> the solution is absolutely right. Unfortunately for <math>n>3</math> the solution is wrong. (Why?)<br />
<br />
--[[User:Drorbn|Drorbn]] 19:46, 24 September 2007 (EDT)<br />
<br />
<br />
===another solution for n<5===<br />
Here's a not-too-sketchy solution for n=3 and a sketch of how the same approach works for n=4. The strategy also works for a 2-legged roach but I really really can't do 5 or more. Perhaps someone else can help!<br />
Note that for n=4 we have a torus, not a sphere! <br />
p.s. Sorry if the image is upside-down or hard to get to, I am just learning how to deal with technology <br />
[[Image:roach.pdf||center||]]</div>Katiemann