I have been reading Fundamentals of Differential Geometry by Serge Lang. Let me briefly describe his definitions. In the following, bilinear forms are always assumed to be continuous.

A Banach space $\mathbf{E}$ (over $\mathbb{R}$) is self dual if there is a symmetric nonsingular bilinear form $\mathbf{E\times E}\to\mathbb{R},(v,w)\mapsto\langle v,w\rangle$, where nonsingular means that the induced map $\mathbf{E}\to L(\mathbf{E},\mathbb{R}),v\mapsto(w\mapsto\langle v,w\rangle)$ is a toplinear isomorphism (i.e., the usual isomorphism of Banach spaces).

Question 1. Is this equivalent to the fact that $\mathbf{E}$ is reflexive?

Then he explains the bijection between the space $L^2_{\text{sym}}(\mathbf{E})$ of symmetric bilinear forms on $\mathbf{E}$ and the space $L(\mathbf{E},\mathbf{E})$, when $\mathbf{E}$ is self dual. They are related by $$\lambda(x,y)=\langle Ax,y\rangle,\quad\lambda\in L^2_{\text{sym}}(\mathbf{E}),\,x,y\in\mathbf{E}.$$ Then $\lambda\leftrightarrow A$ is the intended bijection.

Question 2. I have proved that this is a toplinear isomorphism. Is that correct?

A metric on $\mathbf{E}$ is a nonsingular symmetric bilinear form. When $\mathbf{E}$ is a Hilbert space (self dual via the inner product), a Riemannian metric on $\mathbf{E}$ is a metric for which the corresponding operator $A$ is positive definite, i.e., there is $\varepsilon>0$ such that $\langle Ax,x\rangle\geq\varepsilon\langle x,x\rangle$ for all $x$.

Question 3. For Riemannian metrics, why is $A$ invertible? Also, is the definition equivalent to $\langle Ax,x\rangle\geq0$ for all $x$, as in the finite dimensional case?

Later, when he talks about the functorial behavior of metrics on vector bundles, he uses this: Let $\mathbf{E}'$ be a closed subspace of $\mathbf{E}$ that is complemented, where $\mathbf{E}'$ and $\mathbf{E}$ are self dual. If $\lambda$ is a metric (resp. Riemannian metric) on $\mathbf{E}$, then there is an induced metric (resp. Riemannian metric) on $\mathbf{E}'$ defined by restriction.

Question 4. Why is the restriction of a nonsingular bilinear form also nonsingular? In fact I think this is not true, even in the finite dimensional case...

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    $\begingroup$ One question in one post is a good idea. $\endgroup$ – Kavi Rama Murthy Mar 13 at 9:19
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    $\begingroup$ @KaviRamaMurthy Well, in that case I would have to repeat the definitions for four times... And in fact, the primary reason I'm asking these questions is that I found several mistakes in the book, and it makes me wonder whether I'm the one who's wrong. Finally, I'm not sure whether it's fine to ask a question which merely indicates an error in a specific book, so... $\endgroup$ – Colescu Mar 13 at 10:35

For question 1:

No, definitely not. Reflexive means that the canonical inclusion into the bidual is an isomorphism. Your statement is equivalent to a space being isomorphic to its dual, there exist plenty of reflexive spaces that are not isomorphic to their dual (most of the infinite dimensional ones).

An interesting question is whether or not your statement implies that a space is reflexive. Again this is false. There are spaces isomorphic to their bidual, but for which the canonical inclusion is not an isomorphism. If $J$ is such a space then $J\oplus J^*$ is isomorphic to its own dual but the space itself is not reflexive. (I know these spaces exist in the complex case, but am not so sure for the real case).

For question 2:

Yes, this map should be an isomorphism.

For question 3:

The definition is not equivalent to $\langle A x, x\rangle >0$ for $x\neq0$. Consider the map $\ell^2\to \ell^2$ given by continuous linear extension of $A(e_n)=2^{-n}e_n$. This map is positive but does not have the bounded below property but satisfies the above inequality.

$A$ is invertible because $$\epsilon\|x\|^2=\epsilon \langle x,x\rangle≤\langle A x, x\rangle ≤ \|Ax\|\,\|x\|$$ and $\|Ax\|≥\epsilon\|x\|$, ie $A$ is bounded from below. Since $A$ is positive it is hermitian and thus from bounded below implying the kernel is zero you get that the cokernel must also be zero. The map is then bijective, bounded from below implies open in this setting and you get isomorphism.

For question 4:

As you note the restriction of a non-singular form to a subspace is not necessarily non-singular. This can go wrong in the simplest possible case, namely dimension $2$ to dimension $1$.

The restriction of a positive definite form is however positive definite.


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