L3a. Generalized linear algebra: the symmetries (II).

Here you have the file that Hudson wrote about the symmetries of $V\oplus V^*$. He uses the regular value theorem to prove that $\textrm{O}(V\oplus V^*)$ is a Lie group, computes the Lie algebra (by using the differential of the map he was using) and then looks at the Lie algebra structure. (If anyone wants to add r ask anything, remember that you can comment on the posts or send me files.)



Just for clarity, let me put here the way I was justifying what is the equation describing the Lie algebra $\mathfrak{o}(V\oplus V^*)$. Since $\textrm{O}(V\oplus V^*)$ is a linear Lie group, we can describe its Lie algebra $\mathfrak{o}(V\oplus V^*)$  by taking derivatives of curves inside $\textrm{O}(V\oplus V^*)$ passing by $\textrm{Id}$ at time $0$.
$$\mathfrak{o}(V\oplus V^*) = \{ c'(0) \; : \; c:(-\varepsilon,\varepsilon)\to \textrm{O}(V\oplus V^*), c(0)= \textrm{Id} \}.$$
This will tell us what the condition for the elements in $\mathfrak{o}(V\oplus V^*)$ is.

The condition for the elements $A\in \textrm{O}(V\oplus V^*)$ is $\langle Av,Aw \rangle = \langle v,w \rangle$ for any $v,w \in V\oplus V^*$. Recall that the pairing is $\langle X+\xi, Y+\eta \rangle = \frac{1}{2} (i_X \eta + i_Y \xi )$, so we can write
$$\langle v,w \rangle = v^T  C w,$$
for $N = \left( \begin{array}{cc} 0 & \frac{1}{2} \\ \frac{1}{2} & 0 \end{array} \right)$. In other words, $A\in \textrm{O}(V\oplus V^*)$ if f $A^T N A = N$.

Now, given a curve $c(t)$ in $\textrm{O}(V\oplus V^*)$, it satisfies $c(t)^T N c(t) = N$. Since we are interested in $c'(0)$, we differentiate:
$$c'(t)^T N c(t) + c(t)^T N c'(t) = 0,$$
which at $t=0$, gives $A^T N + NA = 0$.

By looking at $A$ as a block matrix,
$$A =  \left( \begin{array}{cc} E & \beta \\ B & F \end{array} \right),$$
the condition $A^T N + NA = 0$ gives $F = - E^T$, $B+B^T =0$ and $\beta + \beta^T = 0$. This means that an element in $\textrm{O}(V\oplus V^*)$ consists of  $E\in \textrm{End}(V)$ (acting as $-E^T$ on $V^*$ ), $B\in \wedge^2 V^*$ and  $\beta\in\wedge^2 V$.

We thus have
$$\mathfrak{o}(V\oplus V^*) = \textrm{End}(V)\oplus \wedge^2 V^* \oplus \wedge^2 V,$$
where the direct sum refers to vector spaces, not to Lie algebras! (as Hudson showed).