Why $V_i = \mathcal{P}_i (M,g)$?

Question

In the book Le Spectre d'une Variété Riemannienne of Berger, Gauduchon and Mazet, they explain the following :

Let $(M,g)$ a Riemannian manifold, and suppose for all $i \in \mathbb{N}$ a vector subspace $V_i$ of $\mathcal{C}^{\infty}(M)$ in such a way that the two following conditions are verified :

1. For all $i$, there exists $\lambda_i \in \mathbb{R}$ such that for all $\varphi \in V_i$, we get $\Delta \varphi = \lambda_i \varphi$;

2. The sum $\sum_{i \in \mathbb{N}} V_i$ is dense in $\mathcal{C}^{\infty}(M)$ in quadratic mean. Then the spectrum of $(M,g)$ is the set of $\lambda_i$ and, for all $i$, $V_i = \mathcal{P}_i(M,g)$, where $\mathcal{P}_i(M,g)$ is the eigensubspace de $\mathcal{C}^{\infty}(M)$ formed of eigenfunctions related to $\lambda$.

In sentences that followed this lemma, there was a comment. Could someone develop on that following comment? Sometimes when I read some books or article, I notice that the explanations are there, but without many details. Some details in the last paragraph are superfluous. I wish someone could develop these details so that I understand.

It is clear that $V_i \subset \mathcal{P}_i (M,g)$. It follows that $V_i$ is of finite dimension, then closed for the topology of the quadratic mean. Suppose that $V_i \not = \mathcal{P}_i (M,g)$. Then we could find $\phi \in \mathcal{P}_i (M,g)$ orthogonal to $V_i$. Moreover, $\varphi$ would be orthogonal for all $V_j$ for $j \not = i$. In conclusion $\varphi$ would be orthogonal to $V_j$ for all $j$, that couldn't happen. Then $V_i = \mathcal{P}_i (M,g)$

Thanks in advance!

Answer 1

Since $\Delta$ is formally self-adjoint on $C^\infty(M)$, eigenspaces corresponding to distinct eigenvalues are orthogonal to each other, i.e., for all $j \neq i$, $\mathcal{P}_i(M,g) \subset \mathcal{P}_j(M,g)^\perp \cap C^\infty(M)$. Hence, for all $j \neq i$, since $V_j \subset \mathcal{P}_j(M,g)^\perp \cap C^\infty(M)$, it follows that $\mathcal{P}_i(M,g) \subset \mathcal{P}_j(M,g)^\perp \cap C^\infty(M) \subset V_j^\perp \cap C^\infty(M)$.

So, suppose by contradiction that there exists some non-zero $\phi \in \mathcal{P}_i(M,g) \cap V_i^\perp$. On the one hand, by assumption, $\phi \in \mathcal{P}_i(M,g) \cap V_i^\perp \subset V_i^\perp \cap C^\infty(M)$. On the other hand, since $\phi \in \mathcal{P}_i(M,g)$, it follows that for all $j \neq i$, $\phi \in \mathcal{P}_i(M,g) \subset \mathcal{P}_j(M,g)^\perp \subset V_j^\perp \cap C^\infty(M)$. Putting everything together, then, we see that $\phi \in V_j^\perp$ for all $j$, so that $$ \phi \in \cap_j \left(V_j^\perp \cap C^\infty(M)\right) = \left(\sum_j V_j\right)^\perp \cap C^\infty(M) = \{0\}, $$ where $\left(\sum_j V_j\right)^\perp \cap C^\infty(M) = \{0\}$ precisely because $\sum_j V_j$ is dense in $C^\infty(M)$ with respect to the $L^2$ norm (i.e., your "quadratic mean").