Maybe it is a trivial question but I cannot see it. I an working on Riemannian manifolds and I have a bilinear form $Q$ on a conformally flat manifold that for orthogonal vector fields $X,Y$ it verifies $Q(X,Y)=0$.
In the paper I am reading, it proves that this implies that $Q=\lambda g_0$, where $\lambda : M \rightarrow \mathbb{R}$ and $g_0$ a flat metric. In the proof, it says
This is a generality on quadratic forms and follows easily by "polarization".
Why is this?
Best Answer
By linear algebra, there is a linear map $P: T_x\to P_x$ so that $$ \langle Y, P(X)\rangle =Q(X, Y). $$ In fact, under an orthonormal basis of $T_x$, $P$ has the same matrix as $Q$.
Therefore, $\langle Y, P(X)\rangle=0$ for all $Y\perp X$. In another word, $P(X)\perp Y$ for all $Y\in X^\perp$, where $X^\perp\subset T_x$ is the (codimension one) subspace of all vectors perpendicular to $X$. So the vector $P(X)$ is perpendicular to the space $X^\perp$; such vecor must be a multiple of $X$, i.e. $P(x)=\lambda_x X$.
Thus every vector $X$ is a real eigenvector of $P$. Then all the eigenvalues must be the same: when $X, Y$ are independent, from $P(X+Y)=P(X)+P(Y)=\lambda_1 X+\lambda_2Y$ one the one hand, and $P(X+Y)=\lambda_3(X+Y)$ on the other hand we see $\lambda_1=\lambda_2=\lambda_3$. So $P(X)=\lambda X$ for all $X$, thus $$Q(X, Y)= \langle Y, P(X)\rangle=\lambda \langle X, Y\rangle. $$