I have a bilinear form $\phi$ on a complex vector space V and I have to prove the following:

1) if $\phi$ is symmetric non-degenerate it is possible to choose a basis for V such that $\phi$ is described by $X^TY$. Deduce that special linear Lie algebra $\mathfrak{so}(V,\phi)\simeq \mathfrak{so_n}$

where $\mathfrak{so}(V,\phi)= \{A \in M_{n,n}(\mathbb{C}) : (A(u),v)+(u,A(v))=0, \forall u,v \in V\}$

2) if $\phi$ is skew-symmetric non-degenerate it is possible to choose a basis for V such that $\phi$ is described by $X^TJY$, where $$J=\begin{bmatrix} 0 & \mathbb{I}\\ -\mathbb{I} & 0\\\end{bmatrix}$$

I know that the matrix $B$ associated to $\phi$ must have $\det B \neq 0$, but how to relate this with the basis? Why there is a difference in the basis for symmetric and skew-symmetric bilinear forms?

I am struggling with this exercise, can anyone help me?


I try to answer (2). You can try to think about (1) afterwards.

Define $U := \{u \in V: J(u,v) = v \text{ for any } v \in V \}$. Since $J$ is non-degenerate, then $U = \{0\}$. So, if $V$ is not trivial, there exists an element $e_1 \in V \setminus \{0\}$ such that $J(e_1,f_1) \neq 0$ for some $f_1 \in V$. Up to rescaling, you can assume $J(e_1,f_1) = 1$. Call $W := \text{Span}(e_1,f_1)$ and define $$W^J := \{u \in V: J(u,w) = 0 \text{ for any }w \in W\}.$$ Let us take a look at $W \cap W^J$. If $v \in W \cap W^J$, then $v = ae_1+bf_1$ for some $a,b$, and $J(v,e_1) = 0 = J(v,f_1)$. But then $J(v,e_1)=J(ae_1+bf_1,e_1) = -b=0$ and similarly $a=0$. So $W \cap W^J = \{0\}$. Let now $v$ be any vector in $V$. If $J(v,e_1) = -a$ and $J(v,f_1) = b$, then you can write $$v = be_1+af_1+v-be_1-af_1.$$ You have that $be_1+af_1 \in W$ and \begin{align} J(v-be_1-af_1,e_1) & = J(v,e_1)+a = a-a=0\\ J(v-be_1-af_1,f_1) & = J(v,f_1)-b = b-b = 0. \end{align} This tells you that any vector $v \in V$ can be written as a sum of a vector in $W$, that is $be_1+af_1$, and a vector in $W^J$, namely $v-be_1-af_1$. Consequently $V = W \oplus W^J$. If $W^J = \{0\}$, then you are done and $J$ can be written in the basis $\{e_1,f_1\}$ as $$ \left( \begin{matrix} J(e_1,e_1) & J(e_1,f_1) \\ J(f_1,e_1) & J(f_1,f_1) \end{matrix} \right) = \left( \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right). $$ Otherwise, choose $e_2 \neq 0$ in $W^J$ and repeat the process getting $f_2$ such that $J(e_2,f_2)=1$. Going on you will find a basis $\{e_1,e_2,\dots,e_n,f_1,f_2,\dots,f_n\}$ of $V$ such that $$J(e_i,e_j) = 0, \quad J(e_i,f_k) = \delta_{ik}, \quad J(f_i,f_j) = 0,$$ where $\delta_{ij}$ denotes the Kronecker delta. The process ends after $n$ steps, as $\dim V < \infty$. Notice that the presence of $J$ forces $\dim V = 2n$, i.e. the dimension of $V$ is even.

FYI: a non-degenerate skew-symmetric bilinear form $J$ like this is generally called symplectic form on $V$, and $(V,J)$ is then called symplectic space.

  • $\begingroup$ Thanks! Can I ask you another question? Why $(A(u),v)+(u,A(v)) = A^T J+ J A ?$ $\endgroup$ – Phi_24 Oct 2 '18 at 19:56
  • $\begingroup$ Take a look here. Just consider that your $J$ is $\Omega$. $\endgroup$ – Gibbs Oct 2 '18 at 20:02

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