What does $[L]=[I]^{-1}[II]$ mean?

Question

I have a question about one of the equations in my notes.

Matrix representations of Weingarton map, first fundamental form and second fundamental form satisfies $[L]=[I]^{-1}[II]$

According to my understanding, $[II]$ is really the matrix representation of a bilinear map from $T_pM\times T_pM \to R$. And so is $[I]$, so $[I]^{-1}$ represents a map from $R \to T_pM \times T_pM$?

But then $L$ is a map that takes in one vector, and outputs anothe vector. So the left hand side and the right hand side doesn't seem to be the same thing at all. Not sure if I'm confusing myself. Any idea?

Answer 1

No, in this context $[I]^{-1}$ means the matrix inverse of the first fundamental form matrix (e.g. the metric matrix). It doesn't mean the inverse of the first fundamental form (which in any case is not defined: there are a huge number of pairs of vectors whose inner product is $1$, for example.)

Answer 2

Just to complete what user7530 and Travis have written:

$\operatorname{I\!I}$ is a symmetric bilinear form and $\operatorname{I}$ is a nondegenerate symmetric bilinear form. In this situation you have a unique $\operatorname{I}$-self adjoint operator associated to $\operatorname{I\!I}$, i.e. an operator (endomorphism) $L$ such that $$\operatorname{I}(Lx, y) = \operatorname{I}(x, Ly) = \operatorname{I\!I}(x,y)\,.$$ If you take some basis of your vector space $V$ ($= T_pM$) to represent the bilinear forms $\operatorname{I}$ and $\operatorname{I\!I}$ and the endomorphism $L$ by square matrices, then these matrices satisfy the relation that you wrote. It's easy to show directly, but one possible interpretation as pointed out by Travis is that $$L = {\left(\operatorname{I}^\flat\right)}^{-1} \circ \operatorname{I\!I}^{\flat}$$ where $\operatorname{I}^\flat$ is the map $V \to V^*$ defined by $\operatorname{I}^\flat(x) = \operatorname{I}(x , \cdot)$, same for $\operatorname{I\!I}^{\flat}$.