# Are all conjugacy classes in $\text{GL}_n(\mathbb R)$ path-connected?

Suppose $$A$$ and $$B$$ are conjugate invertible real $$n \times n$$-matrices. Does there always exist a path from $$A$$ to $$B$$ inside their conjugacy class?

I thought I had an easy proof for odd $$n$$ which goes as follows, but it was incorrect as pointed out in this answer. To show where my misunderstanding arised, here is the wrong argument.

Suppose there exists a real matrix $$P$$ such that $$B = PAP^{-1}$$. By replacing $$P$$ with $$-P$$ if necessary, we can assume that $$\det P > 0$$ (this is what goes wrong in even dimensions, see this question). Then we have that $$P = e^Q$$ for some real matrix $$Q$$ (since the image of the exponential map is the path-component of the identity in $$\text{GL}_n(\mathbb R)$$). But now the path $$t \mapsto e^{tQ}Ae^{-tQ}$$ is a path connecting $$A$$ to $$PAP^{-1} = B$$.

This edit comes a bit late, but the way I see it is a bit different than the other answers so I'll write it anyway.

Again, I use the example from the question that you link to: $$A$$ is the rotation by $$\frac \pi 2$$ in the euclidean plane, and $$B$$ is the rotation by $$-\frac \pi 2$$.

Now any conjugate of $$A$$ by a matrix with positive determinant will correspond to a linear map $$\varphi$$ such that for any non-zero vector $$v$$, the vectors $$v$$ and $$\varphi (v)$$ in this order make a positive basis of the plane.

Conversely, a conjugate of $$A$$ by a matrix with negative determinant (which is the case of $$B$$) will correspond to a linear map $$\varphi$$ such that for any non-zero vector $$v$$, the vectors $$v$$ and $$\varphi (v)$$ in this order make a negative basis of the plane.

A path from $$A$$ to $$B$$ has to cross the set of matrices that have real eigenvalues, and such a matrix cannot be conjugate to $$A$$.

• I like this geometric approach, this is the solution I will not forget. – Levi Nov 4 '19 at 0:50
• @Levi Thanks :) – Arnaud Mortier Nov 4 '19 at 0:51

We can use the counterexample from your other question to answer this one. Let $$A=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$. Matrices $$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ must have trace $$a+d$$ equal to $$0$$ and determinant $$ad-bc=-a^2-bc$$ equal to $$1$$. This implies that such a matrix cannot have $$b=0$$, since $$-a^2\leq 0$$. Therefore the set of conjugates of $$A$$ is disjoint union of open sets defined by $$b>0$$ and $$b<0$$. Neither of those sets is empty, since one contains $$A$$ and the other contains $$-A$$. Thus the conjugacy class of $$A$$ is disconnected.

• This is much nicer than my argument! (But slightly less detailed, whence I'm leaving mine up.) – darij grinberg Nov 4 '19 at 0:30
• This is an awesome way of seeing this. This particular example also shows that I have been trying to prove the wrong thing when answering the other question. So I am particularly grateful! – Levi Nov 4 '19 at 0:33
• Sorry for unaccepting your answer, but I cannot resist going with coordinate-freeness :) – Levi Nov 4 '19 at 0:50

Other people have given counterexamples, so I would like to demonstrate that counterexamples are somewhat rare. Here is an attempt at a conceptual explanation for both why this is not true in general and also when it is true. First, we note that if we have a path $$P(t)$$ in $$GL_n(\mathbb R)$$ with $$P(0)=P_0$$ and $$P(1)=P_1$$, then $$P(t)AP(t)^{-1}$$ gives us a path inside the conjugacy class of $$A$$. Since $$GL_n(\mathbb R)$$ has two path components (given by sign of determinant), this shows that the conjugacy class of $$A$$ is the union of two path connected subsets (conjugating by things with positive or negative determinant), and it will be path connected if these two subsets intersect.

If $$PAP^{-1}=QAQ^{-1}$$ where $$\det(P)>0$$ and $$\det(Q)<0$$, we have $$A=(P^{-1}Q)A(P^{-1}Q)^{-1}$$, so $$A$$ commutes with a matrix with negative determinant. The converse is also true, so we have the following result.

Lemma: The conjugacy class of $$A$$ is path connected if and only if $$A$$ commutes with some matrix with negative determinant.

This gives several conditions that would ensure conjugacy classes are path connected.

• If $$n$$ is odd, since then $$\det(-I_n)=-1$$
• If $$\det(A)<0$$
• If $$\mathbb R^n$$ as a direct sum of two $$A$$-invariant subspaces where one summand is odd dimensional, (e.g, if the Jordan normal form of $$A$$ has a block of odd size with a real eigenvalue).
• If $$\mathbb R^{n}$$ is the direct sum of two $$A$$-invariant subspaces, and the restriction of $$A$$ to either subspace has negative determinant.

One can check that Jordan blocks only commute with matrices that are upper triangular and have only a single eigenvalue, and if $$n$$ is even and that eigenvalue is real, such a matrix could not have negative determinant. So these give counterexamples.

I suspect one could give a nice characterization of all counterexamples in terms of JNF. However, I have not worked out the details.

No.

Here is a counterexample: Let $$n=2$$, $$A=\left( \begin{array} [c]{cc} 0 & -1\\ 1 & 0 \end{array} \right)$$ and $$B=-A=\left( \begin{array} [c]{cc} 0 & 1\\ -1 & 0 \end{array} \right)$$. Then, $$A$$ and $$B$$ are conjugate, since $$B=C^{-1}AC$$ for the invertible matrix $$C=\left( \begin{array} [c]{cc} 1 & 0\\ 0 & -1 \end{array} \right)$$. Thus, there exists a conjugacy class $$\mathcal{C}$$ in $$\operatorname*{GL}\nolimits_{2}\left( \mathbb{R}\right)$$ that contains both $$A$$ and $$B$$. However, there exists no path from $$A$$ to $$B$$ in $$\mathcal{C}$$. Why not?

Probably there is a nice conceptual reason [EDIT: yes, and @Wojowu explains it in his answer], but you can just as well brute-force it: I claim that every matrix in $$\mathcal{C}$$ has a nonzero $$\left( 1,2\right)$$-th entry. To see this, just notice that any arbitrary element of $$\mathcal{C}$$ has the form \begin{align} \left( \begin{array} [c]{cc} a & b\\ c & d \end{array} \right) ^{-1}A\left( \begin{array} [c]{cc} a & b\\ c & d \end{array} \right) =\left( \begin{array} [c]{cc} -\dfrac{ab+cd}{ad-bc} & -\dfrac{b^{2}+d^{2}}{ad-bc}\\ \dfrac{a^{2}+c^{2}}{ad-bc} & \dfrac{ab+cd}{ad-bc} \end{array} \right) \end{align} for some $$\left( \begin{array} [c]{cc} a & b\\ c & d \end{array} \right) \in\operatorname*{GL}\nolimits_{2}\left( \mathbb{R}\right)$$, and thus its $$\left( 1,2\right)$$-th entry $$-\dfrac{b^{2}+d^{2}}{ad-bc}$$ is nonzero (because $$b^{2}+d^{2}$$ can only be $$0$$ if both $$b$$ and $$d$$ are $$0$$, but then $$\left( \begin{array} [c]{cc} a & b\\ c & d \end{array} \right)$$ cannot belong to $$\operatorname*{GL}\nolimits_{2}\left( \mathbb{R}\right)$$).

Thus, every matrix in $$\mathcal{C}$$ has a nonzero $$\left( 1,2\right)$$-th entry. But if there was a path from $$A$$ to $$B$$ in $$\mathcal{C}$$, then some point on this path would be a matrix in $$\mathcal{C}$$ with zero $$\left( 1,2\right)$$-th entry (since the $$\left( 1,2\right)$$-th entries of $$A$$ and $$B$$ have opposite signs). Thus, there cannot be a path from $$A$$ to $$B$$ in $$\mathcal{C}$$.

• There is also a geometric way to see it if you're interested. – Arnaud Mortier Nov 4 '19 at 0:48
• @ArnaudMortier I'm interested. – Abhimanyu Pallavi Sudhir Nov 29 '19 at 13:22
• @AbhimanyuPallaviSudhir Thanks! You will find it in my answer which is currently the accepted one I think. – Arnaud Mortier Nov 29 '19 at 13:47
• Ah I missed that, thanks. – Abhimanyu Pallavi Sudhir Nov 29 '19 at 15:04

Notice that the assertion "since the image of the exponential map is the path-component of the identity in $$GLn(R)$$" is false.

Firstly, the path-component of the identity in $$GLn(R)$$ is the set of matrices with $$>0$$ determinant. Thus (inside this component) there is a path linking your $$P$$ and $$I$$ (in fact, there is nothing to prove).

Secondly, $$diag(-1,-2)$$ has $$>0$$ determinant and is not in the image of the real matrices by the exponential map.

EDIT.

Let $$SC(A)$$ be the conjugacy class of $$A\in M_n(\mathbb{R})$$.

*In dimension $$2$$, $$SC(A)$$ is non-connected in 2 cases

a. The eigenvalues of $$A$$ are non-zero conjugate complex; then $$SC(A)$$ is homeomorphic to a hyperboloid of two sheets.

b. $$A$$ is non-zero and non-diagonalizable; then $$SC(A)$$ is homeomorphic to a conical surface with the apex cut off.

*In dimension $$4$$, we consider the matrices $$A_1=diag(U,U)$$ where $$U=\begin{pmatrix}0&1\\0&0\end{pmatrix}$$ and $$A_2=diag(V,V)$$ where $$V=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$$. Then, using the Aaron's test, we can prove that $$SC(A_1),SC(A_2)$$ are non-connected (it's easy for $$A_1$$ and more difficult for $$A_2$$).

*Assume that we randomly choose $$A\in M_n(\mathbb{R})$$ -the $$(a_{i,j})$$ follow iid normal laws- We deduce that follows

$$\textbf{Proposition}.$$ When $$n\rightarrow +\infty$$, the probability that $$SC(A)$$ is connected tends to $$1$$.

$$\textbf{Proof}$$. A random matrix $$A$$ has distinct complex eigenvalues with probabiity $$1$$. Then, up to a real change of basis, $$A=diag(a_1I_2+b_1V,\cdots,a_pI_2+b_pV,\lambda_1,\cdots,\lambda_q)$$, where $$2p+q=n$$, the $$(b_i)$$ are non-zero and the $$(\lambda_j)$$ are real distinct.

Thus $$SC(A)$$ is connected iff $$q\not=0$$ (Aaron's test). When $$n$$ tends to $$+\infty$$, the mean of the number of real zeroes of a polynomial of degree $$n$$ is in $$\Omega(\sqrt{n})$$; we can deduce that the probability that $$A$$ has, at least, one real eigenvalue tends to $$1$$ when $$n$$ tends to $$+\infty$$. $$\square$$

• Sorry for not reacting to this sooner. I have now finally managed to convince myself that what you say is true. Thanks for pointing it out! – Levi Nov 29 '19 at 13:09
• This statement also struck me when I first read it but then I focused on answering the question and forgot about it. Thanks for this. – Arnaud Mortier Nov 29 '19 at 13:47
• @Levi . More precisely, the real matrices that can be written in the form $e^Q$ are the squares of real matrices. – loup blanc Nov 29 '19 at 14:02
• @Arnaud Mortier . Thanks Arnaud. I wrote this because I was surprised that no one would react. – loup blanc Nov 29 '19 at 14:05