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).
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).
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