# Are nearby subalgebras of matrix algebras conjugate?

Let $$k=\mathbb{R}$$ or $$\mathbb{C}$$ and let $$A$$ be a finite-dimensional $$k$$-algebra. If $$A$$ is simple, then the Skolem-Noether theorem says that any two algebra homomorphisms $$f, g: A \to M_n(k)$$ are conjugate in $$M_n(k)$$, i.e., there exists a matrix $$X \in M_n(k)$$ such that for all $$a \in A$$, $$f(a) = X g(a) X^{-1}$$.

It we do not assume that $$A$$ is simple, then this result is generally false (see e.g. the example given by Denis Serre at Conjugation between commutative subalgebras of a matrix algebra?).

My question is: Is this result still true if $$f$$ and $$g$$ are "close"? That is, if $$f$$ and $$g$$ are close, then they are conjugate?

Here by "close" I mean that for some small $$\varepsilon>0$$, we have $$\lVert f(a) - g(a)\rVert \leq \varepsilon \lVert a\rVert$$, for some submultiplicative norm on $$A$$.

• Could you be explicit on what you mean by "this result"?
– YCor
Jan 17 at 18:17
• I meant that if $f$ and $g$ are close then they are conjugate. I edited the question accordingly. Jan 17 at 19:09
• I understood this; what's missing is what's the assumption on $A$.
– YCor
Jan 17 at 19:34
• In any case since there are 1-parameter deformations, one should easily find a 1-parameter family of subalgebra depending continuously on a real parameter, that are locally pairwise non-conjugate (and even non-isomorphic). But I'm not sure this would answer your question.
– YCor
Jan 17 at 19:36
• No assumption on $A$, just a finite-dimensional algebra over $k$. Jan 17 at 21:42

I'll describe a 1-parameter family of nonisomorphic 4-dimensional subalgebras of $$M_4(K)$$. Consider, for $$t\in K^*$$, the matrices $$X=\begin{pmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix},\;Y_t=\begin{pmatrix} 0 & 1 & 1 & 0\\ 0 & 0 & 0 & t\\ 0 & 0 & 0 & -t \\ 0 & 0 & 0 & 0 \end{pmatrix},\;Z=\begin{pmatrix} 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}.$$ Then $$XY_t=tZ$$, $$Y_tX=Z$$, and all other pairwise products (including squares) between these matrices are zero. Hence they form the basis of a 3-dimensional (non-unital) subalgebra $$N_t$$. These algebras are pairwise non-isomorphic, except $$N_t\simeq N_{t^{-1}}$$ (indeed, in $$N_t$$, the system of equations $$\{xy=\lambda yx;\quad x^2=y^2=0\}$$ has a solution $$(x,y)$$ with $$xy\neq 0$$ iff $$\lambda\in\{1,t,1/t\}$$).

Let now $$A_t$$ be the 4-dimensional unital subalgebra with basis $$(I,X,Y_t,Z)$$. These are also pairwise nonisomorphic (except $$t\leftrightarrow t^{-1}$$), since $$N_t$$ consists of the set of nilpotent elements in $$A_t$$.

Since they are not isomorphic, they are not conjugate.

Now let $$A$$ be 5-dimensional, with basis $$(I,X,Y,Z,Z')$$ with $$I$$ identity, $$XY=Z'$$, $$YX=Z$$, and all other products (not involving $$I$$) being zero. Let $$f_t:A\to A_t\subset M_4(K)$$ map $$I\mapsto I$$, $$X\mapsto X$$, $$Y\mapsto Y_t$$, $$Z\mapsto Z$$, $$Z'\mapsto tZ$$. Then $$f_t$$ is a surjective homomorphism. However they have non-isomorphic images (up to $$t\leftrightarrow t^{-1}$$). While for $$K$$ being real or complex numbers, for $$t'$$ close enough to $$t$$, the homomorphisms $$f_t$$ and $$f_{t'}$$ are close.

• Now looking at this example more closely, it seems that the equation $xy = \lambda yx$ has in fact a solution for any $\lambda$, just maybe not $x = X$ $y = Y$ or $x= Y$, $y=X$. So I am not sure how to show that the algebras $N_t$ are indeed non-isomorphic. Feb 3 at 22:02
• @MatthiasLudewig you're perfectly right. However, if one requires in addition $x^2=y^2=0$ my statement becomes true. I have corrected accordingly. Thank you for having checked carefully! [Note: actually I had picked this algebra $N_t$ in a list of small-dimensional nilpotent algebras so I knew in advance they were non-isomorphic, so I was not very careful in checking it.]
– YCor
Feb 4 at 16:01
• This makes sense. Thanks for the clarification! Feb 4 at 21:53

Here's another example. It's quite distinct from my previous answer since in my previous answer the homomorphisms $$f_t$$ are non-injective and have images that are not isomorphic (and hence non-conjugate).

Here I provide an example with injective homomorphisms $$f_t$$ (so that their images are isomorphic). I'll not be explicit, using a dimensional argument "there are too many injective homomorphisms for all of them to be conjugate".

Namely, $$A=A_n$$ is the unital (commutative associative) algebra of dimension $$n+1$$, with basis $$(I,X_1,\dots,X_n)$$ with $$I$$ unit and all other products being zero. Hence, a representation of $$A$$ is just the data of $$n$$ matrices squaring to zero and with pairwise zero product. And I'll assume $$n=8$$ although passing to $$n\ge 8$$ should be immediate (while decreasing to smaller $$n$$ should be possible, although I'm not really sure about $$n=2,3$$). Let $$R_n$$ be the nilradical of $$A_n$$, that is (with basis $$(X_1,\dots,X_n)$$). So a representation of the unital algebra $$A_n$$ is the same as a representation of the (non-unital) algebra $$R_n$$.

Consider in $$M_8(K)$$ the matrices by $$4+4$$ blocks $$\begin{pmatrix}0 & *\\ 0 & 0\end{pmatrix}$$. These form a 16-dimensional subspace of $$M_8(K)$$. Its 8-Grassmanian has dimension 64. Each element in this 8-Grassmanian corresponds to a faithful representation of $$R_8$$, and hence of $$A_8$$. But the conjugacy action is that of $$\mathrm{PGL}_8(K)$$ and $$\mathrm{PGL}_8$$ has dimension 63. This implies, for $$K$$ being $$\mathbf{R}$$ or $$\mathbf{C}$$, that the conjugation action cannot be locally transitive. Hence every algebra in this family has arbitrary close neighbors in the same family, which are not conjugate to it.

• Thanks for this additional example. I think I know what I am getting it, but could you elaborate on what you mean by the "8-Grassmannian"? Jan 18 at 10:07
• @MatthiasLudewig In a vector space of dimension $n$, by $k$-Grassmanian I mean the set of $k$-dimensional subspaces (which for $0\le k\le n$ forms a variety of dimension $k(n-k)$).
– YCor
Jan 18 at 10:40

$$\DeclareMathOperator\ker{ker}$$Assume that $$A$$ is semisimple, so by Wedderburn's theorem, $$A$$ is isomorphic to $$M_{n_1}(k)\oplus M_{n_2}(k)\oplus\dots\oplus M_{n_r}(k)$$ where $$n_1\leq n_2\leq\dots\leq n_r$$ positive integers. Thus, two homomorphisms $$f,g:A\to M_n(k)$$ are equivalent if and only if $$\ker{f}=\ker{g}$$.

Claim: Given a norm $$\|.\|_A$$ on $$A$$, there exists $$\varepsilon>0$$ such that for any two homomorphisms $$f,g:A\to M_n(k)$$, $$\ker{f}=\ker{g}$$ whenever $$\|f(a)-g(a)\|\leq \varepsilon\|a\|_A$$ for all $$a\in A$$.

Proof. Let $$p_i$$ denote the identity of $$M_{n_i}(k)$$ so that $$\{p_1,\dots,p_r\}$$ is a set of pairwise orthogonal projections such that $$p_1+\dots+p_r=1$$. Let $$\varepsilon = \frac{1}{2}\min\Big\{\frac{1}{\|p_1\|_A},\dots,\frac{1}{\|p_r\|_A}\Big\}.$$

Now suppose $$\ker{f}\neq \ker{g}$$. Then, either $$p_i\subseteq \ker{f}\backslash \ker{g}$$ for some $$i=1,\dots,r$$, or $$p_i\subseteq \ker{g}\backslash \ker{f}$$ for some $$i=1,\dots,r$$. In either case, $$\|f(p_i)-g(p_i)\|=1>\varepsilon\|p_i\|_A$$.

Edit/Correction: The original claim was incorrect. Rather than deleting the answer, we modified it to get some benefit, a partial answer out of what was written. Please see Yves de Cornulier's reply for a counterexample for non-semisimple $$A$$.

• The answer of Denis Serre that is linked to in the question seems to contradict your assertion that "two homomorphisms $f,g : A \to M_n(k)$ are equivalent if and only if $\ker(f)=\ker(g)$", because in his answer one has $\ker(f)=\ker(g)=\{0\}$. Jan 18 at 1:46
• I blundered. This assertion is valid given $f$ and $g$ are irreducible representations, but incorrect for any two homomorphisms. Jan 18 at 2:30