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Apologies if this is a stupid question ; it is at least a natural question. Let $V$ be a finite dimensional space over $\mathbb R$ or $\mathbb C$. Denote by ${\mathcal L}(V)$ the vector space of all endomorphisms of $V$ ; it has dimension $n^2$, where $n={\sf dim}(V)$. Also, denote by ${\cal B}(V)$ the space of all bilinear maps ${\mathcal L}(V) \times {\mathcal L}(V) \to {\mathcal L}(V)$ (this has dimension $n^6$). Say that a $b\in {\cal B}(V\times V,V)$ is invariant if it satisfies

$$ b({g}^{-1}f_1g,{g}^{-1}f_2g)={g}^{-1}b(f_1,f_2)g $$ for any invertible $g\in{\mathcal L}(V)$ and any $f_1,f_2 \in {\mathcal L}(V)$. Clearly the invariant bilinear maps form a subspace of ${\cal B}(V)$, let us denote it by ${\cal I}(V)$.

For $n=2$, I computed a basis of ${\cal I}(V)$ and found that ${\cal I}(V)$ has dimension $7$ (see update below).

In general, is the dimension of ${\cal I}(V)$ known? Or even better, a simple description of the elements of ${\cal I}(V)$, or of some basis of it ?

Update 10/15/2002 As requested in a comment, more details when $n=2$ : in this case ${\cal I}(V)$ has a basis made up of six “simply described” elements and a complicated seventh one :

$$ \begin{array}{ccl} b_1(f,g)&=&fg \\ b_2(f,g)&=&gf \\ b_3(f,g)&=&{\sf tr}(f)g \\ b_4(f,g)&=&{\sf tr}(g)f \\ b_5(f,g)&=&{\sf tr}(f){\sf tr}(g){\mathbf{id}}_V \\ b_6(f,g)&=& {\sf tr}(fg) {\mathbf{id}}_V \end{array} $$

and the seventh one can be defined in terms of matrices :

$$ b_7\Bigg( \left(\begin{matrix} x_{11} & x_{12} \\ x_{21} & x_{22} \\ \end{matrix}\right), \left(\begin{matrix} y_{11} & y_{12} \\ y_{21} & y_{22} \\ \end{matrix}\right) \Bigg)=\left(\begin{matrix} x_{21}y_{12}-x_{12}y_{21} & (y_{11}-y_{22})x_{12}-(x_{11}-x_{22})y_{12} \\ -(y_{11}-y_{22})x_{21}+(x_{11}-x_{22})y_{21} & -x_{21}y_{12}+x_{12}y_{21} \\ \end{matrix}\right) $$

Edit 24/10/2012 : Some mildly incorrect statements corrected above.

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Absolutely not a stupid question.. Can you help us with the computed case of $n=2$? –  Berci Oct 14 '12 at 14:06
    
@Berci : maybe not stupid, but definitely the kind of thing one expects to be studied already (hence the reference-request tag). I’ll post the details on $n=2$ when I have the time. The $n=3$ case should be computable also. –  Ewan Delanoy Oct 14 '12 at 14:48
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up vote 1 down vote accepted

The link provided by Mariano Suarez-Alvarez online answers most of what can be asked on this theme.

Here, I’ll try to provide an elementary proof for the specific question in the OP. The result is that for $n\geq 2$, ${\cal I}(V)$ always has dimension six and has a basis $b_1,b_2, \ldots b_6$ as described in the update in the OP (so $b_7$ is exceptional and only exists in dimension $2$).

Now fix a basis $(v_1,v_2, \ldots ,v_n)$ of $V$. Let $({v_1}^{*},{v_2}^{*}, \ldots ,{v_n}^{*})$ denote the dual basis in $V^*$, and put $e_{ij}(x)={v_i}^{*}(x)e_j$. Then the $e_{ij}$ form a basis for ${\cal L}(V)$, which we call $\cal E$.

We will now use the invariance property, making some special endomorphisms act on $V$ and hence on ${\cal L}(V) \otimes {\cal L}(V)$ (let us denote this space by ${\cal T}(V)$ ; then ${\cal E}\otimes{\cal E}=\lbrace e \otimes f | e,f \in {\cal E} \rbrace$ is a basis of this space).

Let $p_1,p_2, \ldots ,p_r$ be distinct primes (so that no nontrivial multiplicative relation exists between them), and let $d$ be the diagonal endormorphism defined by $d(v_i)=p_iv_i$. Then $d$ also acts diagonally on ${\cal L}(V)$ and ${\cal T}(V)$ : we have

$$ \begin{array}{c} d^{-1}e_{ij}d=\frac{p_j}{p_i}e_{ij}, \\ (d^{-1}e_{ij}d) \otimes (d^{-1}e_{kl}d)=\frac{p_jp_l}{p_ip_k}(e_{ij} \otimes e_{kl}) \end{array} $$

The eigenspaces for the action of $d$ on ${\cal L}(V)$ are (from now on $i,j,k,l$ etc always denote different indices ) :

For the eigenvalue $1$, the $n$-dimensional subspace $V_1={\sf Vect}(e_{xx})_{1 \leq x \leq n}$

For the eigenvalue $\frac{p_j}{p_i}$, the $1$-dimensional subspace $V_2(i,j)={\sf Vet}(e_{ij})$.

The eigenspaces for the action of $d$ on ${\cal T}(V)$ are :

For the eigenvalue $1$, the $n^2$-dimensional subspace $W_1$.

For the eigenvalue $\frac{p_j}{p_i}$, the $4(n-1)$-dimensional subspace $W_2(i,j)$

For the eigenvalue $\frac{p_j^2}{p_i^2}$ the one-dimensional subspace $W_3(i,j)$.

For the eigenvalue $\frac{p_jp_k}{p_i^2}$ the two-dimensional subspace $W_4(i,j,k)$.

For the eigenvalue $\frac{p_k^2}{p_ip_j}$ the two-dimensional subspace $W_5(i,j,k)$.

For the eigenvalue $\frac{p_jp_l}{p_ip_k}$ the four-dimensional subspace $W_6(i,j,k,l)$, where

$$ \begin{array}{lcl} W_1 &=& {\sf Vect}(e_{xx} \otimes e_{xx})_{1 \leq x \leq n} \oplus {\sf Vect}(e_{xx} \otimes e_{yy},\ e_{xy} \otimes e_{yx})_{1 \leq x \neq y \leq n} \\ W_2(i,j) &=& W_{2,1}(i,j) \oplus W_{2,2}(i,j) \\ & & W_{2,1} (i,j)={\sf Vect}(e_{ii} \otimes e_{ij},e_{ij} \otimes e_{ii},e_{ij} \otimes e_{jj},e_{jj} \otimes e_{ij}) \\ & & W_{2,2} (i,j)={\sf Vect}(e_{ij} \otimes e_{xx},e_{xx} \otimes e_{ij},e_{ix} \otimes e_{xj},e_{xj} \otimes e_{ix}, )_{x \in [1…n] \setminus \lbrace i,j \rbrace} \\ W_3(i,j) &=& {\sf Vect}(e_{ij} \otimes e_{ij}) \\ W_4(i,j,k) &=& {\sf Vect}(e_{ij} \otimes e_{ik},e_{ik} \otimes e_{ij}) \\ W_5(i,j,k) &=& {\sf Vect}(e_{ik} \otimes e_{jk},e_{jk} \otimes e_{ik}) \\ W_6(i,j,k,l) &=& {\sf Vect}(e_{ij} \otimes e_{kl},e_{il} \otimes e_{kj},e_{kj} \otimes e_{il},e_{kl} \otimes e_{ij}) \\ \end{array} $$

By hypothesis, $b$ commutes with the action of $d$, so

$$ (1) \ b(W_1) \subseteq V_1, \ b(W_2(i,j)) \subseteq V_2(i,j),\ b=0 \ \text{on all the} \ W_3,W_4,W_5,W_6. $$

This yields a special constraint on the form of each of the $n^4$ individual maps $f(e_{t_1t_2} \otimes e_{t_3t_4})$. Those constraints can be refined further by considering the following : for $A\subseteq {\cal L}(V)$ let $S(A)$ denote the stabilizer of $A$ in $L(V)$, i.e.

$$ S(A)=\Bigg\lbrace s \in L(V) \Bigg| \ \forall a\in A, \ \ sa=as \Bigg\rbrace $$

Then any $s\in S(\lbrace e_{t_1t_2},e_{t_3t_4} \rbrace)$ fixes the element $e_{t_1t_2} \otimes e_{t_3t_4}$. We deduce that not only $f(e_{t_1t_2} \otimes e_{t_3t_4})$ must inhabit the space assigned to him by (1), it must also be in the set $ S'(\lbrace e_{t_1t_2},e_{t_3t_4} \rbrace) $ of endomorphisms fixed by all the invertible elements of $S(\lbrace e_{t_1t_2},e_{t_3t_4} \rbrace)$. So we need to compute $S(\lbrace e_{t_1t_2},e_{t_3t_4} \rbrace)$ and $S'(\lbrace e_{t_1t_2},e_{t_3t_4} \rbrace)$ in general : a straightforward set of linear algebra computations shows that

$$ \begin{array}{lcl} S(e_{ii}) &=& {\sf Vect}(e_{ii}) \oplus {\sf Vect} (e_{xy})_{x\neq i,y \neq i} \\ S(e_{ij}) &=& {\sf Vect}(e_{ii}+e_{jj}) \oplus {\sf Vect} (e_{xy})_{x\neq j,y \neq i} \\ S(\lbrace e_{ii},e_{jj} \rbrace) &=& {\sf Vect}(e_{ii},e_{jj}) \oplus {\sf Vect} (e_{xy})_{x\not\in \lbrace i,j \rbrace, y \not\in \lbrace i,j \rbrace } \\ S(\lbrace e_{ii},e_{ij} \rbrace) &=& {\sf Vect}(e_{ii}+e_{jj}) \oplus {\sf Vect} (e_{xy})_{x\neq i, y \not\in \lbrace i,j \rbrace } \\ S(\lbrace e_{ii},e_{ji} \rbrace) &=& {\sf Vect}(e_{ii}+e_{jj}) \oplus {\sf Vect} (e_{xy})_{x\not\in \lbrace i,j \rbrace , y\neq i} \\ S(\lbrace e_{ij},e_{ji} \rbrace) &=& {\sf Vect}(e_{ii}+e_{jj}) \oplus {\sf Vect} (e_{xy})_{x\not\in \lbrace i,j \rbrace , y\not\in \lbrace i,j \rbrace } \\ S(\lbrace e_{ij},e_{kk} \rbrace) &=& {\sf Vect}(e_{ii}+e_{jj},e_{kk}) \oplus {\sf Vect} (e_{xy})_{x\not\in \lbrace j,k \rbrace , y\not\in \lbrace i,k \rbrace } \\ S(\lbrace e_{ij},e_{jk} \rbrace) &=& {\sf Vect}(e_{ii}+e_{jj}+e_{kk} ) \oplus {\sf Vect} (e_{xy})_{x\not\in \lbrace i,j \rbrace , y\neq k} \\ \end{array} $$

and hence

$$ \begin{array}{lcl} S'(e_{ii}) &=& {\sf Vect}(e_{ii},{\sf id}_V) \\ S'(e_{ij}) &=& {\sf Vect}(e_{ii},e_{jj},e_{ij},{\sf id}_V) \\ S'(\lbrace e_{ii},e_{jj} \rbrace) &=& {\sf Vect}(e_{ii},e_{jj},{\sf id}_V) \\ S'(\lbrace e_{ii},e_{ij} \rbrace) &=& {\sf Vect}(e_{ii},e_{ij},e_{ji},e_{jj},{\sf id}_V) \\ S'(\lbrace e_{ii},e_{ji} \rbrace) &=& {\sf Vect}(e_{ii},e_{ij},e_{ji},e_{jj},{\sf id}_V) \\ S'(\lbrace e_{ij},e_{ji} \rbrace) &=& {\sf Vect}(e_{ii},e_{ij},e_{ji},e_{jj},{\sf id}_V) \\ S'(\lbrace e_{ij},e_{kk} \rbrace) &=& {\sf Vect}(e_{ii}+e_{jj},e_{ij},e_{kk},{\sf id}_V) \\ S'(\lbrace e_{ij},e_{jk} \rbrace) &=& {\sf Vect}(e_{jj},e_{ij},e_{ik},e_{jk},{\sf id}_V) \\ \end{array} $$

Combining the two constraints, we obtain the following : there are structure constants $a_1,a_2, \ldots, a_{16}$ such that for any distinct $i,j,k$

$$ \begin{array}{lcl} b(e_{ii} \otimes e_{ii}) &=& a_1e_{ii}+a_2{\sf id}_V \\ b(e_{ii} \otimes e_{jj}) &=& a_3e_{ii}+a_4e_{jj}+a_5{\sf id}_V \\ b(e_{ij} \otimes e_{ji}) &=& a_6e_{ii}+a_7e_{jj}+a_{8}{\sf id}_V \\ b(e_{ii} \otimes e_{ij}) &=& a_{9}e_{ij} \\ b(e_{ij} \otimes e_{ii}) &=& a_{10}e_{ij} \\ b(e_{ij} \otimes e_{jj}) &=& a_{11}e_{ij} \\ b(e_{jj} \otimes e_{ij}) &=& a_{12}e_{ij} \\ b(e_{ij} \otimes e_{kk}) &=& a_{13}e_{ij} \\ b(e_{kk} \otimes e_{ij}) &=& a_{14}e_{ij} \\ b(e_{ij} \otimes e_{jk}) &=& a_{15}e_{ik}\\ b(e_{jk} \otimes e_{ij}) &=& a_{16}e_{ik} \\ b(e \otimes f) &=& 0 \ \text{for all the other } \ e \otimes f \ \text{in the canonical basis.} \end{array} $$

( those structure constants should depend on the indices $i,j,k$ a priori, but the invariance by permutation matrices allows us to see they do not).

If we write out the invariance with respect to the action of the transvection $p={\sf id}_V+te_{12}$, we see that we can express linearly $a_{8},a_9,a_{10},a_{11},a_{12}$ in terms of $a_1,a_2 \ldots a_7$ (I omit some computational details here).

In dimension $2$, this is sufficient since we already have seven independent maps ($b_1$ to $b_7$ as in the OP).

In dimension $n \geq 3$, similarly we can write out the invariance with respect to endomorphisms of the form ${\sf id}_V+ue_{12}+ve_{13}+we_{23}$, and this way we can express all the $a_i$ linearly in terms of $a_1,a_2, \ldots ,a_6$ only.

Since we already have six independent solutions ($b_1$ to $b_6$ as in the OP), this concludes the proof.

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