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I have searched but could not find the information.

It is well known that

$$ \exp \mathfrak{gl} \to GL^+ $$

is only surjective if $\mathfrak{gl}$ are the complex-valued matrices. If they are real valued, the map is not surjective.

But then from the Baker–Campbell–Hausdorff formula1, we know that there exists many matrices such that $e^Xe^Y=e^Z$.

Can we say that exponentials of matrices are a group under multiplication? Because $\forall X \forall Y: e^Xe^Y=e^Z$, as per the formula above?

Rodrigo de Azevedo
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Anon21
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    Even if we ignore the "sufficiently small" hypothesis, that equation only says that exponentials of matrices are closed under multiplication—a group has other hypotheses that must be checked. – Greg Martin Feb 17 '23 at 17:01
  • @GregMartin The inverse of $e^X$ is $e^{-X}$. The identity is $e^0=1$. And it is associative $(e^Xe^Y)e^A=e^X(e^Ye^A)$. – Anon21 Feb 17 '23 at 17:03
  • @GregMartin The question remaining is "does the multiplication of two sufficiently small matrices always produce a sufficiently small matrice". – Anon21 Feb 17 '23 at 17:08
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    As a side note, why in the world would you include a screenshot of Wikipedia in your question? It's Wikipedia! You can link to it! You can copy and paste! Either one of those would have been more useful. – Misha Lavrov Feb 17 '23 at 17:11
  • Please do not rely on pictures of text. – Shaun Feb 17 '23 at 17:23
  • It's hard to imagine that "sufficiently small" is closed under multiplication. If $X$ is sufficiently small, is $nX$ sufficiently small for all positive integers $n$? (since $(e^X)^n = e^{nX}$) – Greg Martin Feb 17 '23 at 17:40
  • @GregMartin Here https://en.wikipedia.org/wiki/Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula in the section "Existence results", it seems to suggest that the exponentials of matrices form a grouplike if they are sufficiently small, and elements of a grouplike are closed under multiplication. But I cannot parse the text with certainty, because there are too many unfamiliar concepts to me. Can you kindly check and share your opinion? – Anon21 Feb 17 '23 at 17:57
  • @DinosaurEgg Quite interesting. I guess I have my work cut out. Any thing else you want to chip in would be appreciated. – Anon21 Feb 17 '23 at 18:57
  • @DinosaurEgg I don't suppose I could beg you for a fetched out proof? Please. – Anon21 Feb 17 '23 at 19:23
  • @DinosaurEgg I added a bounty, if you would like to have a go. – Anon21 Feb 19 '23 at 17:14
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    The statement you are after is simply false. My suggestion is to examine the image of the space of real 2x2 matrices under $\exp$ and check that it is not a group. – Moishe Kohan Feb 20 '23 at 01:39
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    Apparently, a (real) matrix has a logarithm if and only if it has an inverse and a square root. See one, two, three. – mr_e_man Feb 21 '23 at 04:27

2 Answers2

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Following the comment of @Moishe Kohan:

Given a $2\times2$ matrix $M$, let $t=\text{tr}(M)$ be the trace of $M$, let $F=\text{tf}(M)$ be the trace-free part of $M$, so $M=\tfrac t2I+F$, and let $d=\det(F)$ be the determinant of $F$. A straightforward calculation (or Cayley-Hamilton) gives $F^2=-d\,I$, so the exponential is

$$\exp(M)=\exp(\tfrac t2)\exp(F)=\exp(\tfrac t2)\cdot\begin{cases}\cos(\sqrt d)\,I+\frac{1}{\sqrt d}\sin(\sqrt d)\,F,\quad d>0\\I+F,\quad d=0\\\cosh(\sqrt{-d})\,I+\frac{1}{\sqrt{-d}}\sinh(\sqrt{-d})\,F,\quad d<0\end{cases}$$

$$=M'=\tfrac{t'}2I+F'$$

The trace of this is

$$t'=\text{tr}(\exp(M))=\exp(\tfrac t2)\cdot\begin{cases}2\cos(\sqrt d),\quad d>0\\2,\quad d=0\\2\cosh(\sqrt{-d}),\quad d<0\end{cases}$$

and the trace-free part is

$$F'=\text{tf}(\exp(M))=\exp(\tfrac t2)\cdot\begin{cases}\frac{1}{\sqrt d}\sin(\sqrt d)\,F,\quad d>0\\F,\quad d=0\\\frac{1}{\sqrt{-d}}\sinh(\sqrt{-d})\,F,\quad d<0\end{cases}$$

which has determinant

$$d'=\det(F')=\exp(t)\cdot\begin{cases}\sin^2(\sqrt d),\quad d>0\\0,\quad d=0\\-\sinh^2(\sqrt{-d}),\quad d<0\end{cases}$$

The important part is that if $d'<0$ (or $d'=0$ and $F'\neq0$) then $t'>0$.

We can take any matrix $M$ with $d\leq0$, and multiply the two exponentials $\exp(M)=M'$ and $\exp\left(\begin{bmatrix}0&-\pi\\\pi&0\end{bmatrix}\right)=-I$, to get another matrix $M''=-M'$. This has $d''=d'\leq0$ and $F''=-F'$ but $t''=-t'<0$, so $M''$ cannot be a single exponential.

For example:

$$M''=\exp\left(\begin{bmatrix}0&-\pi\\\pi&0\end{bmatrix}\right)\exp\left(\begin{bmatrix}1&0\\0&-1\end{bmatrix}\right)$$

$$=\begin{bmatrix}-e&0\\0&-1/e\end{bmatrix}$$

$$t''=\text{tr}(M'')=-e-1/e<0$$

$$F''=\text{tf}(M'')=\frac{-e+1/e}{2}\begin{bmatrix}1&0\\0&-1\end{bmatrix}$$

$$d''=\det(F'')=-\left(\frac{-e+1/e}{2}\right)^2<0$$

Thus, the image of $\exp$ is not a group.

mr_e_man
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This is not true. For example, $$\exp \begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix} =\begin{bmatrix} 1&1 \\ 0&1 \end{bmatrix} \ \text{and}\ \exp \begin{bmatrix} 0&\pi \\ -\pi&0 \end{bmatrix} =\begin{bmatrix} -1&0 \\ 0&-1 \end{bmatrix}.$$ However, $$\exp \begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix}\exp \begin{bmatrix} 0&\pi \\ -\pi&0 \end{bmatrix} =\begin{bmatrix} 1&1 \\ 0&1 \end{bmatrix}\begin{bmatrix} -1&0 \\ 0&-1 \end{bmatrix} = \begin{bmatrix} -1&-1 \\ 0&-1 \end{bmatrix}$$ is not the exponential of any real matrix.

The point is that the BCH formula gives a formula for $\exp(X) \exp(Y)$ as the exponential of a formal sum, but that sum may or may not be convergent, and the particular matrices in my example are outside the radius of convergence.

See the Wikipedia article on the BCH formula.

Alex Selby
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David E Speyer
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  • "... is not the exponential of any real matrix." - Why? – mr_e_man Feb 21 '23 at 22:02
  • In the context of my answer, the reason would be that it has negative trace ($t'<0$, hence $d>0$), and its trace-free part is non-zero but has zero determinant ($d'=0$, hence $\sin^2(\sqrt d)=0$), which is impossible for an exponential (as $\sin(\sqrt d)=0$ implies $F'=0$, a contradiction). – mr_e_man Feb 21 '23 at 22:14
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    The usual way I see this verified is: If $e^X$ were this matrix, then the generalized eigenvalues of $X$ would have to in $(\pi i) + (2 \pi i) \mathbb{Z}$. Since $X$ is real, the set of its generalized eigenvalues would be closed under complex conjugation, so they must be $\pm (2 k+i) (\pi i)$ for some $k$. But then $X$ would have distinct complex eigenvalues, so it would be complex diagonalizable, so $e^X$ would be complex diagonalizable, contradiction. – David E Speyer Feb 21 '23 at 23:45
  • I'm sure there are many other ways to do this. – David E Speyer Feb 21 '23 at 23:45