Let $V$ be a real finite-dimensional vector space of dimension $n$. Let $End_{\Bbb{R}}V$ be the real vector space of linear mappings from $V$ to $V$. $End_{\Bbb{R}}V$ has dimension $n^2$.
The trace is the unique (up to scale) $Tr\in(End_{\Bbb{R}}V)^{*}$ such that $Tr(A\circ B)=Tr(B\circ A)$ for every $A,B\in End_{\Bbb{R}}V$. Normally we take the one that verifies $Tr(Id)=n$, where $Id$ is the identity endomorphism of $V$. This defines a non-degenerate symmetric bilinear form: $B_{Tr}:End_{\Bbb{R}}V\times End_{\Bbb{R}}V\to \Bbb{R}$, $B_{Tr}(A,B):=Tr(A\circ B)$.
It is a well known result that non-degenerate symmetric bilinear forms over a real finite-dimensional vector space are classified by the index and the signature, so I wonder:
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Is there a formula to calculate the index and the signture of $B_{Tr}$ (maybe in terms of the dimension)? In other words, how do I find the corresponding $\pm1s$? There are $n^2$ of them.
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Who is an orthonormal basis for $(End_{\Bbb{R}}V,B_{Tr})$? In order to solve this, does it help the fact that the orthogonal complement of the symmetric endomorphisms are the alternating ones and viceversa?
For the answerers: I have no knowledge about algebraic geometry. If you are able to connect your answer with ideas of the field of differential geometry it will be appreciated (I'm teaching myself it and that's how I came up with this question). Though coordinates help to understand things, finish with free-coordinate arguments will be appreciated too. Still any help will be welcome. Thanks in advance.
Best Answer
Once you choose some inner product $g$ on $V$, you have the notion of an an adjoint map. The bilinear form $$ B_{g}(T,S) = \operatorname{tr} \left( T^{*} \circ S \right) $$ depends on $g$ (via the adjoint) and defines an inner product (i.e a non-degenerate symmetric positive definite bilinear form) on $\operatorname{End}(V)$ which is sometimes called the "standard" inner product on $\operatorname{End}(V)$ and is commonly used in differential geometry. Now, any operator $T \colon V \rightarrow V$ can be decomposed uniquely as $T = T_1 + T_2$ where $T_1$ is self-adjoint ($T_1^{*} = T_1$) and $T_2$ is skew-adjoint $(T_2^{*} = -T_2$). Explicitly, we have $$ T = \underbrace{\frac{1}{2} \left( T + T^{*} \right)}_{T_1} + \underbrace{\frac{1}{2} \left( T - T^{*} \right)}_{T_2}. $$ Hence $\operatorname{End}(V) = \operatorname{End}_{\textrm{self}}(V) \oplus \operatorname{End}_{\textrm{skew}}(V)$. Note that if $T_1 \in \operatorname{End}_{\textrm{self}}(V), T_2 \in \operatorname{End}_{\textrm{skew}}(V)$, we have $$ B_{\operatorname{tr}} \left( T_1, T_2 \right) = \operatorname{tr} \left( T_1 \circ T_2 \right) = \operatorname{tr} \left( \left( T_1 \circ T_2 \right)^{*} \right) = -\operatorname{tr} \left( T_2 \circ T_1 \right) = -\operatorname{tr} \left( T_1 \circ T_2 \right) = -B_{\operatorname{tr}} \left( T_1, T_2 \right) \implies \\ B_{\operatorname{tr}} (T_1,T_2) = 0.$$ This shows that the decomposition of $\operatorname{End}(V)$ is $B_{\operatorname{tr}}$-orthogonal so it is enough to compute the signature of $B_{\operatorname{tr}}$ restricted to each of the subspaces. Note that $$ B_{\operatorname{tr}}|_{\operatorname{End}_{\textrm{self}}(V)} = B_g|_{\operatorname{End}_{\textrm{self}}(V)}, \qquad B_{\operatorname{tr}}|_{\operatorname{End}_{\textrm{skew}}(V)} = -B_g|_{\operatorname{End}_{\textrm{skew}}(V)} $$ so $B_{\operatorname{tr}}$ is positive definite on $\operatorname{End}_{\textrm{self}}(V)$ and negative definite on $\operatorname{End}_{\textrm{skew}}(V)$ which shows that the index of $B_{\operatorname{tr}}$ is $\left( \frac{n(n+1)}{2}, \frac{n(n-1)}{2} \right)$.
This calculation of the signature of $B_{\operatorname{tr}}$ works by choosing an "auxiliary" inner product $g$ on $V$ and then comparing $B_{\operatorname{tr}}$ with $B_g$. By choosing different inner products, you get different forms $B_g$ and different direct sum decompositions of $\operatorname{End}(V)$ but the final result is the same.