Let $\mathfrak s$ be the set of symmetric $n$ by $n$ matrices, and let $S$ be the set of positive definite matrices. I'm trying to understand why the exponential map $f : \mathfrak s \rightarrow S$, $f(X) = \exp(X)$ is a local diffeomorphism. This question has been asked before but I didn't understand the answer.

What we have to show is that for every $X \in \mathfrak s$, the tangent space map at $X$

$$df_X: T_X(\mathfrak s) \rightarrow T_{\exp(X)}(S)$$

is an invertible linear transformation. In the answer, I don't understand the identifications being made and the notation, so I am asking the question again here.

It is first of all not clear to me why the manifolds $\mathfrak s$ and $S$ have the same dimension, nor how we should think of the two tangent spaces.

Since $\mathfrak s$ is a Euclidean space, we could identify $T_X(\mathfrak s)$ with $\mathfrak s$ itself. I don't understand how to make sense out of $T_{\exp(X)}(S)$ though.


1 Answer 1


First of all, for dimensionality, you should be able to convince yourself that $\dim \mathfrak{s} = \frac{n(n+1)}{2}$. This can be done easily through counting the number of independent parameters in a symmetric matrix. To prove that $S$ is a manifold, one can either show that $S$ is an open subset of $\mathfrak{s}$ or that $S_n \approx GL_n(\mathbb{R})/O(n)$ using the homogeneous space construction theorem. This also proves that $\dim S = \dim GL_n(\mathbb{R}) - \dim O(n) = n^2 - \frac{n(n-1)}{2} = \frac{n(n+1)}{2}$. Therefore they are of the same dimension.

To further show how $S$ and $\mathfrak{s}$ are related, you should recognize that $\mathfrak{s} = T_IS$ since if $\gamma:(-\epsilon, \epsilon)\to GL_n(\mathbb{R})$ with $\gamma(0) = I$ and $\gamma'(0) = X$, then $\eta(t) = \gamma(t)\gamma^T(t)$ defines a smooth curve in $S$. Then, differentiating we find that

\begin{eqnarray*} \dot{\eta}(0) & = & \dot{\gamma}(0)\gamma^T(0) + \gamma(0)\dot{\gamma}^T(0) \\ & = & X + X^T \end{eqnarray*}

which is symmetric, proving that $T_IS \subseteq \mathfrak{s}$. Then by dimensionality equality holds.

What's remarkable is that the exponential map $\exp:\mathfrak{s} \to S$ is actually a global diffeomorphism. We can see this actually just through the eigendecomposition of symmetric matrices. If $A \in \mathfrak{s}$ then $A = PDP^T$ where $D$ is diagonal with arbitrary entries, and $P$ is orthogonal. Then basic properties of the Taylor series of $\exp$ show that

$$ \exp(A) \;\; =\;\; P\exp(D)P^T $$

where $\exp(D)$ will just have the diagonal entries $e^{d_i}$ which will necessarily be positive and therefore force $\exp(A) \in S$. Similarly, the matrix logarithm, defined by the Taylor series

$$ \log(B) \;\; =\;\; \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} (I - B)^k $$

serves as an inverse to $\exp$ since the logarithm carries the same computational property: $\log(B) = \log(QEQ^T) = Q\log(E)Q^T$ with the diagonal entries of $\log(E)$ just being $\log(\lambda_i)$ where $\lambda_i$ are the diagonal entries of $E$.

This answer should hopefully give you some insight into the relationship between symmetric matrices and the symmetric positive definite matrices, however in your prompt you seem to be further curious about the pushforward of the exponential. In other words, while $\exp:\mathfrak{s} \to S$ is a diffeomorphism, we still have the pushforward map

$$ d\exp_X: T_X\mathfrak{s} \to T_{\exp(X)}S. $$

Computing the purshforward is a much more involved computation, however the fact that $\exp$ is a global diffeomorphism should convince you that $d\exp_X$ is an isomorphism between the tangent spaces of the two manifolds.

  • $\begingroup$ Thanks for your answer. Using the homogeneous space construction theorem, how can we show that $S_n \cong \operatorname{GL}_n(\mathbb R)/O_n$? (granting that the product map $S_n \times O_n \rightarrow \operatorname{GL}_n(\mathbb R)$ is a smooth bijection)? $\endgroup$
    – D_S
    Apr 15, 2020 at 2:20
  • $\begingroup$ Let $GL_n(\mathbb{R})$ act on $S_n$ by $A\cdot P = APA^T$. It is a linear algebra exercise to show that this action is transitive, hence $S_n \approx GL_n(\mathbb{R})/Stab(\cdot)$ where you can pick the stabilizer of any point in $S_n$. If you pick the identity matrix, you can show that $Stab(I) = O(n)$. $\endgroup$
    – Mnifldz
    Apr 15, 2020 at 2:22
  • $\begingroup$ The transitivity of the action is equivalent to the statement of Graham-Schmidt orthogonalization, right? $\endgroup$
    – D_S
    Apr 15, 2020 at 2:46
  • $\begingroup$ Also, are you sure the logarithm converges to a smooth function for all $X \in S_n$? $\endgroup$
    – D_S
    Apr 15, 2020 at 2:51
  • $\begingroup$ One might be able to use Gram-Schmidt, but another argument is the following: any matrix $P \in S$ can be taken back to the identity via some $A \in GL_n(\mathbb{R})$. Given the orthogonal decomposition $P = SES^T$, take $A = E^{-1/2}S^T$, then $A\cdot P = I$. Then it's clear that $P = A^{-1}\cdot I$. Then $P$ can be taken to $K = QFQ^T$ via the matrix $QF^{1/2}E^{-1/2}S^T$. $\endgroup$
    – Mnifldz
    Apr 15, 2020 at 2:54

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