# Search results

• ...(A)+\frac13 \int_\mathbb{R^3}Tr(g^{-1} d g \wedge g^{-1} d g \wedge g^{-1} d g)$$$CS(A^g)=\int_\mathbb{R^3} Tr(A^g \wedge d A^g + \frac23 A^g \wedge A^g \wedge A^g)$$ 4 KB (938 words) - 20:15, 24 August 2018 • ...= A\wedge \mathrm{d}A[/itex] is invariant under $A\mapsto A + \mathrm{d}f$''' \Psi(A + \mathrm{d}f) &= (A + \mathrm{d}f)\wedge \mathrm{d}(A + \mathrm{d}f)\\ 360 bytes (64 words) - 14:51, 18 July 2018 • More on pushforwards, $d^{-1}$, and $d^\ast$. 67 bytes (12 words) - 14:43, 7 February 2014 • $$d(gs) = (dg)s+gds \implies (dg)s = d(gs)-gds$$ D_{g^{-1} A g+g^{-1} d g} (s)\\ 351 bytes (86 words) - 00:15, 23 August 2018 • A^{(gh)} &= (gh)^{-1}A(gh) + (gh)^{-1}\mathrm{d}(gh) \\ &= (gh)^{-1}A(gh) + (gh)^{-1}\Big((\mathrm{d}g)h + g(\mathrm{d}h)\Big)\\ 527 bytes (111 words) - 18:02, 26 July 2018 • ...to produce a $k+1$-form and that $\mathrm{d}\circ \mathrm{d} =0$. ...rm so that $\mathrm{d}f \in \Omega^1(M)$. Thus $d: \mathrm{d} \Omega^0(\mathbb{R}^3) \rightarrow \Omega^1(\mathbb{R}^3)$ is the gr 2 KB (415 words) - 00:42, 28 June 2018 • Fourth red relation on the right should be $D=\partial(d)A^{-1}GA$ 179 bytes (35 words) - 13:00, 8 May 2015 • ...in D}(\Phi_{\vec{e}}^\ast\omega_3)_{21}\prod_{\text{black}\atop \vec{e}\in D}(\Phi_{\vec{e}}^\ast\omega_1)_{10}[/itex]. I'll try to explain and make a p 486 bytes (88 words) - 14:54, 19 March 2014 • Some analysis of $d^{-1}$. 37 bytes (7 words) - 12:28, 31 January 2014 • The correct definition of $d^*$. 43 bytes (7 words) - 13:46, 12 March 2014 • A Drinfel'd-Kohno theorem. 26 bytes (3 words) - 16:50, 13 November 2015 • A naive way to define $d^*$. 39 bytes (8 words) - 13:43, 12 March 2014 • ...a metric space. Prove that the metric itself, regarded as a function $d\colon X\times X\to{\mathbb R}$, is continuous. ...(x,A):=\inf_{y\in A}d(x,y)[/itex], is a continuous function and that $d(x,A)=0$ iff $x\in\bar{A}$. 2 KB (303 words) - 16:50, 25 November 2016 • The dual of $H^0(\mathcal{D}_n)$. 44 bytes (9 words) - 13:40, 12 March 2014 • ...um of all$\mathcal{D}^{pb}_m$, and define$\mathcal{A}^{pb}$as$\mathcal{D}^{pb}/\mathcal{I}$, where$\mathcal{I}\$ is an ideal generated by relations
• I have noticed that the d$x$ always comes before the integrand. Any reason for this or it
• ...w becomes a single variable minimum/maximum problem. We set $\frac{d}{d\epsilon}f(\epsilon)\mid_{\epsilon=0} = 0$, and solve for $x_c</m [itex]\frac{d}{d\epsilon}f(\epsilon)\mid_{\epsilon=0} = \int_0^T dt(\dot{x}_c\dot{x}_q - x_q 2 KB (308 words) - 18:26, 14 May 2018 • ...a metric space. Prove that the metric itself, regarded as a function [itex]d\colon X\times X\to{\mathbb R}$, is continuous. ...(x,A):=\inf_{y\in A}d(x,y)[/itex], is a continuous function and that $d(x,A)=0$ iff $x\in\bar{A}$.