Notes for AKT-140224/0:22:08: Difference between revisions

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We will check that the prospective Lie algebra $g = \mathds{F}^2$ satisfies the Jacobi identity $[x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0$, and thus is indeed a Lie algebra. In contrast with the notation given in lecture, let $\alpha$ and $\beta$ be the basis elements of $\mathds{F}^2$ over $\mathds{F}$, so that $\mathds{F}^2 = \{a\alpha + b\beta: a, b \in \mathds{F}\}$. Now, let $x = a_1\alpha + a_2\beta$, $y = b_1\alpha + b_2\beta$, and $z = c_1\alpha + c_2\beta$, arbitrary elements in $\mathds{F}^2$.
We will check that the prospective Lie algebra $g = F^2$ satisfies the Jacobi identity $[x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0$, and thus is indeed a Lie algebra. In contrast with the notation given in lecture, let $\alpha$ and $\beta$ be the basis elements of $F^2$ over $F$, so that $F^2 = \{a\alpha + b\beta: a, b \in F\}$. Now, let $x = a_1\alpha + a_2\beta$, $y = b_1\alpha + b_2\beta$, and $z = c_1\alpha + c_2\beta$ be arbitrary elements in $F^2$.


Using bilinearity,
Using bilinearity,
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$[y,z] = b_1c_1[\alpha,\alpha] + b_1c_2[\alpha, \beta] + b_2c_1[\beta,\alpha] + b_2c_2[\beta,\beta]$
$[y,z] = b_1c_1[\alpha,\alpha] + b_1c_2[\alpha, \beta] + b_2c_1[\beta,\alpha] + b_2c_2[\beta,\beta]$


Since $[\alpha, \alpha] = [\beta,\beta] = 0$, $[\alpha, \beta] = \alpha$, and $[\beta,\alpha] = -\alpha$, this evaluates to $$[y,z] = (b_1c_2 - b_2c_1)\alpha$$. Similar calculations yield $$[z,x] = (c_1a_2 - c_2a_1)\alpha$$ and $$[x,y] = (a_1b_2 - a_2b_1)\alpha$$. Thus, $[x,[y,z]] + [y,[z,x]] + [z,[x,y]] = (a_2(b_1c_2 - b_2c_1) + b_2(c_1a_2-c_2a_1) + c_2(a_1b_2+a_2b_1))[\beta, \alpha] = -\alpha(a_2b_1c_2 - a_2b_2c_1 + a_2b_2c_1 - a_1b_2c_2 + a_1b_2c_2 - a_2b_1c_2) = 0$, and the Jacobi identity holds.
Since $[\alpha, \alpha] = [\beta,\beta] = 0$, $[\alpha, \beta] = \alpha$, and $[\beta,\alpha] = -\alpha$, this evaluates to $$[y,z] = (b_1c_2 - b_2c_1)\alpha$$ Similar calculations yield $$[z,x] = (c_1a_2 - c_2a_1)\alpha$$ and $$[x,y] = (a_1b_2 - a_2b_1)\alpha$$ Thus, $$[x,[y,z]] + [y,[z,x]] + [z,[x,y]] = (a_2(b_1c_2 - b_2c_1) + b_2(c_1a_2-c_2a_1) + c_2(a_1b_2+a_2b_1))[\beta, \alpha]$$
$$= -\alpha(a_2b_1c_2 - a_2b_2c_1 + a_2b_2c_1 - a_1b_2c_2 + a_1b_2c_2 - a_2b_1c_2) = 0$$

As a result, the Jacobi identity holds.

Latest revision as of 21:54, 12 June 2018

We will check that the prospective Lie algebra $g = F^2$ satisfies the Jacobi identity $[x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0$, and thus is indeed a Lie algebra. In contrast with the notation given in lecture, let $\alpha$ and $\beta$ be the basis elements of $F^2$ over $F$, so that $F^2 = \{a\alpha + b\beta: a, b \in F\}$. Now, let $x = a_1\alpha + a_2\beta$, $y = b_1\alpha + b_2\beta$, and $z = c_1\alpha + c_2\beta$ be arbitrary elements in $F^2$.

Using bilinearity,

$[y,z] = b_1c_1[\alpha,\alpha] + b_1c_2[\alpha, \beta] + b_2c_1[\beta,\alpha] + b_2c_2[\beta,\beta]$

Since $[\alpha, \alpha] = [\beta,\beta] = 0$, $[\alpha, \beta] = \alpha$, and $[\beta,\alpha] = -\alpha$, this evaluates to $$[y,z] = (b_1c_2 - b_2c_1)\alpha$$ Similar calculations yield $$[z,x] = (c_1a_2 - c_2a_1)\alpha$$ and $$[x,y] = (a_1b_2 - a_2b_1)\alpha$$ Thus, $$[x,[y,z]] + [y,[z,x]] + [z,[x,y]] = (a_2(b_1c_2 - b_2c_1) + b_2(c_1a_2-c_2a_1) + c_2(a_1b_2+a_2b_1))[\beta, \alpha]$$ $$= -\alpha(a_2b_1c_2 - a_2b_2c_1 + a_2b_2c_1 - a_1b_2c_2 + a_1b_2c_2 - a_2b_1c_2) = 0$$

As a result, the Jacobi identity holds.

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