Tangent Space of SymSL(n,$\mathbb{R}$) at arbitrary point

I am looking for the tangent space of $SymSL(n,\mathbb{R}) = \{A\in\mathbb{R}^{n\times n} | \,A^{\rm T} = A,\; \det{A}=1 \}$ (actually $n=3$) at an arbitrary $M\in SymSL(n,\mathbb{R})$.

(Actually I am interested in $SymSL^+(n,\mathbb{R}) = \{A\in\mathbb{R}^{n\times n} | \,A^{\rm T} = A,\; \det{A}=1,\; A \text{ is positive definite} \}$, but I have the feeling it boils down to the title of this question.)

Related questions:

Tangent Space of SL(n,ℝ) at arbitrary point, e.g. not at $\mathbb{1}$

Tangent space of Sym(n,ℝ)

The first question's answer doesn't work here because $Sym(n)$ is not a group, the second question's answer is not applicable because $SL(n)$ is not a vector space. Hence $SymSL(n,\mathbb{R})$ is neither a group nor a vector space.

I would very much appreciate any comments.

edit #2: $SL(n,\mathbb{R})$ is connected and smooth. So $SymSL(n,\mathbb{R})$ is the intersection of a connected smooth manifold with a vector space. This makes me believe we can talk about a tangent space in this case. Some more possibly usefull facts: the set $SymSL^+(n,\mathbb{R})$ I am actually interested in is a connected, simply connected and complete Riemannian manifold.

• Does it matter that $SymSL(n,\mathbb{R})$ isn't a group here? – Santana Afton Apr 13 '17 at 7:21
• Yes, the first related question's answer makes use of SL being a group. Then you can just move the tangent space from the identity element to any other by group action. – Everyday Astronaut Apr 13 '17 at 7:23
• So, what about $SymSL(n,\mathbb{R})$? It isn't a group. – Santana Afton Apr 13 '17 at 7:25
• Thats why I don't have an answer to the question of this topic. I also emphasized this fact in the edit. In case you wonder about the solvability of this problem, see the second edit (will be there in 1 min). – Everyday Astronaut Apr 13 '17 at 7:44

Let $\mathcal{S}$ be the vector space (of dimension $k=n(n+1)/2$) of the real symmetric matrices. We consider the algebraic set $V=\{X\in\mathcal{S}| \det(X)=1\}$. Let $A\in V$. Then the tangent space of $V$ in $A$ is $T_A(V)=\{H\in \mathcal{S}|trace(HA^{-1})=0\}$.

Proposition. $V$ has dimension $k-1$.

Proof. This is equivalent to show that $dim(T_A(V))=k-1$ or that the linear form $f:H\in\mathcal{S}\rightarrow trace(HA^{-1})$ is not identically zero; this last point is clear because $f(A^{-1})=trace(A^{-2})>0$.

Remark. Note that $W=\{X\in V|X>0\}$ is open in $V$. Then, the tangent space of $W$ in $A\in W$ is the same as the tangent space of $V$ in $A$. Of course, $dim(W)=k-1$.

EDIT. Answer to @ Dr Doolittle . Let $k\leq n$, $f:\mathbb{R}^n\rightarrow \mathbb{R}^k\in C^1$ , $V=\{x\in\mathbb{R}^n|f(x)=0\}$ and $a\in V$. If $Df_a$ has full rank, then, in a neighborhood of $a$, $V$ is a variety of dimension $n-k$ and the tangent space of $V$ in $a$ is $T_a(V)=\{h\in\mathbb{R}^n|Df_a(h)=0\}$.

Here $f(X)=\det(X)-1$ and $Df_A(H)=trace(HA^{-1})$.

• I would be grateful if you could prove that $T_A(V)$ is the tangent space of $V$ in $A$, or give a reference to a proof. Then the question would be wholly answered. – Everyday Astronaut May 4 '17 at 8:29
• I see it now. Actually this answer has been given in the opening post of the related question math.stackexchange.com/questions/2231893/… with a different notation. Thank you! – Everyday Astronaut May 4 '17 at 9:26
• Addendum: the proof of $T_A(V)$ being the tangent space of $V$ in $A$ simply by deriving the formula $\mathrm{trace}(HA^{-1})=0$ does not necessitate that $T_A(V) \subset S$, i.e. one may think the tangents could be non-symmetric. However they must be symmetric because $V$ is an algebraic variety in the vector space $S$, so all tangent objects must come from $S$. – Everyday Astronaut Jun 20 '17 at 9:32

The following is a rather trivial fact that seems to immediately provide a partial solution. Furthermore, this partial solution may be in fact a complete one in many cases.

Let $N$ be a smooth manifold, and let $L,M\subset N$ be submanifolds. Let $p\in L\cap M$ such that around $p$ the intersection $L\cap M$ is smooth. Then$$T_p(L\cap M)\subset \left(T_pL\right)\cap\left(T_pM\right).$$If $L$ and $M$ intersect transversally, the above $\subset$ is an equality.

Edit: A bit more on transversality: We say $L$ and $M$ intersect each other transversally if for every $p\in L\cap M$ we have$$T_pN=T_pL+T_pM.$$The most important fact about transversality is that whenever $L$ and $M$ intersect each other transversally, it automatically follows that $L\cap M$ is a smooth submanifold. Additionally, we also have $$T_p(L\cap M)=(T_pL)\cap(T_pM)$$for any $p\in L\cap M$. In particular, we know the dimension of $L\cap M$.

• Thank you for this contribution. What are criteria for the transversality of intersections? If not generally, then in the case $L=Sym(n)$, $M=SL(n)$, $N=\mathbb{R}^{n \times n}$. – Everyday Astronaut Apr 25 '17 at 11:05

I am interested in seeing an answer to this question, I haven't been able to get an answer but this is what I do have to help prompt a conclusive answer.

First as noted in the OP the space of positive symmetric matrices is a Riemannian manifold with tangent space at a particular point $\mathbf{P}$ given by $\{ \mathbf{P} \} \times \mbox{Sym}(n,\mathbb{R})$. We are interested in the variety defined by $\{ \mathbf{P} \; : \; \mbox{det}(\mathbf{P}) = 1 \}$, using the point in Amitai's answer leads to $$T_{\mathbf{P}} \mathcal{M} \subset \mbox{Sym}(n,\mathbf{R}) \cap \{ \mathbf{M} \; : \; \mbox{tr}(M) = 0\}.$$ Proving one way or another that subset is proper or not I haven't been able to do, but here is a motivating example in $\mathbb{R}^{2\times 2}$, I also define the symmetric, trace zero matrices $$\mathbf{E}_1 = \begin{bmatrix} 1 & 0\\0 &-1 \end{bmatrix}, \qquad \mathbf{E}_2 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}.$$ Now we want $$0 = \frac{d}{dt}\mbox{det}\,\mathbf{A}(t) = \mbox{tr}\left(\mbox{adj}\, \mathbf{A}(t) \frac{d A}{dt} \right),$$ or $$\mbox{adj} \, A(t) \frac{dA}{dt} = c_1 \mathbf{E}_1 + c_2\mathbf{E}_2,$$ for some constants $c_1$ and $c_2$, although I guess $c_1, c_2$ could be time dependent also? Anyway working with this I get \begin{align} \frac{dA}{dt} = \frac{1}{\mbox{det}(\mbox{A}) } \mbox{A} \left(c_1\mathbf{E}_1 + c_2 \mathbf{E}_2 \right). \end{align} In particular we should have $\mathbf{P} = \mathbf{A}(0)$ and therefore we should be able to find constants $c_1$ and $c_2$ such that $$\mathbf{P} \left(c_1 \mathbf{E}_1 + c_2 \mathbf{E}_2 \right) = \mathbf{E}_i, \qquad i = 1,2$$ a particular instance of this when $i=1$ implies $$\begin{bmatrix} c_1 & c_2 \\ c_2 & -c_1 \end{bmatrix} = \begin{bmatrix} p^{-1}_{11} & -p^{-1}_{12} \\ p^{-1}_{12} & -p^{-1}_{22} \end{bmatrix}$$ (where by $p_{ij}^{-1}$ I mean the $i,j$th entry of the inverse matrix) which jointly seem to suggest that $\mathbf{P} = \mathbf{I}$. On the otherhand keeping $\mathbf{P}$ fixed then $c_2$ is determined as soon as $c_1$ is which suggests the tangent space is only one dimensional.

Would be very interested in comments or a solution for the general case.

• I'm unable to follow the argument from "Now we want". Why do we want that? The only way I can think of the trace coming into play is if $\mathbf{A}(0)=\mathbf{P}=\mathbf{I}$ because only there the derivative of the $\mathrm{det}$ function is the trace. – Everyday Astronaut May 4 '17 at 8:17