## Non-conjugate matrices that are conjugate

Sylvain Cappell was here in Madison today talking about classification of group actions on manifolds, and he mentioned a crazy fact, due to himself and Shaneson, that I didn’t know.  You can have two matrices, i.e. linear transformations of R^d, which are not conjugate in GL_d(R), but they are conjugate by some homeomorphism of R^d!  That’s cool.  And to me kind of amazing!  People who know the paper are welcome to explain in comments why I should not have been amazed by this fact, if that’s indeed the case.

Update:  As a couple of people point out in comments, this fact is really not so amazing as I stated it, because e.g. multiplication by 5 and multiplication by 125 on R^1 are conjugate by x -> x^3!  Looking at Cappell and Shaneson’s “Non-linear similarity,” I see that the real point is to prove this for matrices with eigenvalues on the unit circle.  This makes the eigenvalues “feel” much more like topological invariants, in the sense that there are certain sequences of powers of the matrix that move closer and closer to the identity.  It’s a theorem of Poincare (I learn from C&S) that 2×2 matrices with norm-1 eigenvalues are topologically conjugate if and only if they’re linearly conjugate.  And de Rham showed (at least for orthogonal matrices) that topological conjugacy can’t change any eigenvalue that’s not a root of unity.  Then Nicolaas Kuiper and my UW colleague Joel Robbin extended this to the general linear group, and conjectured that topological conjugacy implied conjugacy in general.  Lots of cases of the Kuiper-Robbin conjecture were proved, e.g. by Sullivan and Schwartz; for instance it is true for matrices of odd prime power order.  So what Cappell and Shaneson did was construct the first counterexamples to the Kuiper-Robbin conjecture.  And more:  they go a long way towards classification of linear transformations with norm-1 eigenvalues up to topological conjugacy, showing e.g. that the two notions agree in dimensions up to 5.

By the way, as Tom Graber pointed out in comment, you can’t make two non-conjugate linear maps conjugate via a diffeomorphism, because you can read the eigenvalues off the action on the tangent space at 0.  But Capell and Shaneson show that you can get the job done with a homeomorphism that’s smooth everywhere except o!  So the obvious obstruction is in some sense the only one.

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## 7 thoughts on “Non-conjugate matrices that are conjugate”

1. Terence Tao says:

This is not quite the same thing, but it surprised me to learn that in many cases the most “efficient” way to solve a linear equation such as $Lu=f$ (where “efficient” means something like “the norm of $u$ is bounded by a constant multiple of the norm of $f$“) is to make $u$ depend nonlinearly on $f$. In particular, while the open mapping theorem for Banach spaces says that efficient solutions always exist when $L$ is a surjective map between Banach spaces, there are examples in which $L$ has no bounded linear right inverse. And these are not just “pathological” examples: for instance, a basic theorem in harmonic analysis is that every function of bounded mean oscillation on the real line can be written as the sum of a bounded function and the Hilbert transform of a bounded function, but it turns out that this decomposition cannot be performed in a bounded linear fashion (though it can be performed in a bounded nonlinear fashion, or else in an unbounded linear fashion).

(This and some other examples of this phenomenon are discussed at this blog post of mine: http://terrytao.wordpress.com/2009/02/01/245b-notes-9-the-baire-category-theorem-and-its-banach-space-consequences/ )

2. NDE says:

Isn’t that already true for d=1? The homeomorphism taking x to x^3 conjugates the matrix [3] to the matrix [27]. The same trick works for any positive [a] and [b] as long as they’re both >1 or both <1.

3. Maybe I am misunderstanding something, but this seems almost trivial.

The group of homeomorphisms is (much!) larger than the group of general linear transformations. So, naturally, many more things are conjugate under the larger group.

One property of conjugating by GL_d(R) is that it preserves the trace. Obviously, that’s not true of conjugating by a homeomorphism.

So we can (again, assuming I have not misunderstood) construct an example even for d=1.

The map x↦x^3 is a homeomorphism of R to itself. Conjugation by GL_1(R) is trivial. But conjugating the matrix “5”, by x↦x^3, yields the matrix “125”.

4. Tom Graber says:

That example is pretty simple, but it still seems remarkable to me, since for diffeomorphisms you’d get the same conjugacy classes, since you could read everything off from the derivative at 0. (I am trying to explain why you _should_ have been amazed. Possibly a fool’s errand…)

5. JSE says:

The point made by NDE and Distler certainly makes it seem non-amazing! I have presumably misinterpreted something so let me check what it is. The relevant paper is “Non-linear similarity,” by Capell and Shaneson, I think.

6. JSE says:

OK, post updated….

7. Andrew Bridy says:

Eric Bach and I wrote a paper on a similar problem for finite fields. There you don’t have any notion of topological conjugacy, but it’s interesting to ask how many conjugacy classes of nxn matrices there are under any permutation whatsoever of (F_q)^n. In this setting there’s a nice way to construct lots of non-linearly conjugate conjugacies: take the companion matrices of distinct degree n irreducible polynomials over F_q whose roots have the same multiplicative order in F_{q^n}. These matrices have distinct characteristic polynomials, so they’re not similar, but you can show that they are conjugate!

The paper is here if you’re interested: http://arxiv.org/abs/1208.3229