# Transversality in Morse theory, linear algebra version

I am working on a product in Morse-Bott homology which has led me to the following considerations and unanswered question. I would be very grateful if anyone could help.

Suppose $H:\mathbb{R}^n \to \mathbb{R}^n$ is a linear map which i symmetric when viewed as a matrix. The spectral theorem gives a decomposition $\mathbb{R}^n\cong E^+ \oplus E^- \oplus E^0$, where $E^{\pm}$ denotes the sum of positive (+) respectively negative $(-)$ eigenspaces of $H$ and $E^0$ denotes the kernel of $H$.

Let $\mathcal{M}$ denote the set of symmetric, positive definite matrices on $\mathbb{R}^n$. I am interested in the $n\times n$ matrix $g^{-1}H$ for $g\in \mathcal{M}$ (corresponding to the Hessian in Morse theory). Denote by $X_g:\mathbb{R}\to \mathbb{R}^{n\times n}$ the path of matrices \begin{align} \mathbb{R}&\to \mathbb{R}^{n\times n} \\ t &\mapsto e^{g^{-1}Ht}. \end{align}

Now define the stable subspace $V^s(g)$ of the operator $X_g$ by \begin{align} V^s(g):=\{v\in \mathbb{R}^n \ |\ X_g(t)v \to 0 \ \text{for} \ t\to \infty \}. \end{align} My question is: How big is the set $\mathcal{S}:=\{ V^s(g)\ |\ g\in \mathcal{M}\}$ in the Grassmannian of $\mathbb{R}^n$?

For example, is it true that any subspace of maximal dimension sitting inside the cone generated by $E^+$ in $\mathbb{R}^n\cong E^+ \oplus E^- \oplus E^0$ is an element of $\mathcal{S}$?

Any help is much appreciated!!! Thanks in advance,

Consider $H$, $g$ as symmetric bilinear forms, not as endomorphisms. Then $g^{-1}H$ is an endomorphism that describes the gradient vector field. Note that $\ker H=\ker g^{-1}H$ is a well-defined subspace. Moreover, the dimensions $n_\pm$ of the positive/negative eigenspaces are determined by $H$ alone.
Every pair of subspaces $V^-$, $V^+$ of dimension $n_-$, $n_+$ such that $\pm H|_{V^\pm}>0$ simultaneously occur as stable and unstable subspace $V^s(g)$, $V^u(g)$ for some $g\in\mathcal M$: Define $g$ such that these spaces are orthogonal to each other and to $\ker H$. On $V^\pm$, put $g=\pm H^{-1}$, on $\ker H$ choose any metric you like.
Let $p,q,r$ be the signature of $H$. Since $g^{-1}$ is symmetric $>0$, $g^{-1}H$ has $p$ positive, $q$ negative $r$ zero eigenvalues and is diagonalizable according to the decomposition $\mathbb{R}^n=Eg^+\oplus E_g^-\oplus E_g^0$. Thus $V^s(g)=E_g^-$ and $dim(V_s^g)=q$. It remains to see if $\mathcal{S}$ contains all vector-spaces of dimension $q$.
• Yes I completely agree. Thanks for your input... Just in case you are curious I have found out that $\mathcal{S}=C$ where $C$ denotes the cone spanned by eigenvectors of negative eigenvalue of $H$. Supposedly there is a proof somewhere in Katok and Hasselblatt's "Introduction to the Modern Theory of Dynamical Systems". – MBIS Aug 25 '15 at 8:11