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Let $K$ be a field with $char(K) \neq 2$. Let $0 \neq V$ be a $K$-vectorspace with $dim(V)< \infty$ and let $B: V \times V \to K$ be a nondegenerate bilinearform. Consider the subspace $W = \{f \in End(V) : B(f(x),y))+B(x,f(y))=0$ for all $x,y \in V \} \subset End(V)$.
My question is if there are always $f,g \in W$ with $f \circ g \neq 0$ (an idempotent $f \neq 0$ would therefore also be enough). This property seems so general that it should be obvious , but except for $0$ I don't know how a single element of $W$ looks like.

korn55
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  • Have you tried writing this down with matrices? For a basic example, choose the standard "scalar product" bilinear form on $V=K^n$, i.e. $B(\pmatrix{x_1\ \vdots \x_n}, \pmatrix{y_1\ \vdots \y_n}) = \sum x_iy_i$. Then remember that for this product, you have $B(Av,w)=B(v, A^Tw)$. Then $W$ consists of which matrices? – Torsten Schoeneberg Apr 30 '20 at 21:08
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    That being said, your title suggests you already know that your $W$ is the Lie algebra $\mathfrak{so}_B$ with respect to the bilinear form (for the standard BLF above, or over an algebraically closed field, that's just $\mathfrak{so}_n$ up to isomorphism). Compare https://math.stackexchange.com/q/3487953/96384 and https://math.stackexchange.com/a/2785625/96384. Then it's an easy check for $n=2$, and for $n=3$ follows just from the fact that these Lie algebras are not abelian – Torsten Schoeneberg Apr 30 '20 at 21:34
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    @TorstenSchoeneberg regarding your first comment, if I choose the standard scalar product for $B$ then I believe $W$ would consist of the skew-symmetric matrices. If S denotes the matrix corresponing to an arbitrary nondegenerate bilinearform $B$, then I believe $W$ would be the set of all matrices $X$ such that $SX+X^TS=0$. But to me this still isn't very descriptive, I just translated from the language of linear maps to the language of matrices. – korn55 May 01 '20 at 00:10
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    You're right! Well, I think the crucial benefit is that composition of endomorphisms translates to matrix multiplication. I'm sure you can find two skew-symmetric matrices whose product is not $0$. – Torsten Schoeneberg May 01 '20 at 00:42
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    @TorstenSchoeneberg Hm, If I take $X$ to be the skew-symmetric matrix having only $1$'s above the diagonal and only $-1$'s below the diagonal, then $X$ is invertible, so $X^2$ can't be $0$. So this should do for the case $S = I_n$, but I don't see how to generalize this for arbitrary (invertible) $S$. – korn55 May 01 '20 at 00:56
  • You're right, the general case is maybe not straightforward. I would apply "big theorems", like scalar extension to an algebraic closure, where the thing becomes isomorphic to the standard $\mathfrak{so}_n$, and then use the method from my second comment. It's entirely possible that there is a much more elementary argument though. – Torsten Schoeneberg May 01 '20 at 19:04

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I assume we are talking about a symmetric bilinear form $B$ here.

Let $n := \mathrm{dim}(V)$. For $n=1$, the space $W$ is reduced to $\{0\}$, but for $n \ge 2$, indeed there exist $f \in W$ with $f\circ f \neq 0$. Much stronger statements can be made with using scalar extension and classification of semisimple Lie algebras (at least if $\mathrm{char}(K)=0$), but here is a rather elementary proof for the claim:

First, it is well known that one can "diagonalise" the symmetric bilinear form (reference for linear algebra books that teach reverse Hermite method for symmetric matrices, Bilinear Form Diagonalisation, How to diagonalize $f(x,y,z)=xy+yz+xz$). That means there is a basis $v_1, ..., v_n$ of $V$ such that $B(v_i, v_j) =0$ for $i\neq j$ and further, if the form is non-degenerate, that all $a_i :=B(v_i, v_i)\neq 0$.

Now define $f\in \mathrm{End}(V)$ as follows:

$$f(v_i):= \begin{cases} v_2 \qquad \; \;\text{ if } i=1\\ \frac{-a_2}{a_1}v_1 \quad \text{ if } i=2 \\0 \qquad \quad\text{ if } i \ge 3\end{cases}$$

Check that $f$ is in $W$, and obviously $(f\circ f) (v_1) = -\frac{a_2}{a_1} v_1 \neq 0$.

Note that in the special case that $B$ is the standard "scalar product" and $v_i$ an orthonormal basis, $W$ consists of all skew-symmetric matrices and the above $f$ corresponds to the matrix

$$\pmatrix{0 &-1& &\\\ 1&0& &\\ &&&\Large{0}\\}$$

which was my motivating example for the general construction.

Torsten Schoeneberg
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