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 $A$.

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    $\begingroup$ Could you be explicit on what you mean by "this result"? $\endgroup$
    – YCor
    Jan 17 at 18:17
  • $\begingroup$ I meant that if $f$ and $g$ are close then they are conjugate. I edited the question accordingly. $\endgroup$ Jan 17 at 19:09
  • $\begingroup$ I understood this; what's missing is what's the assumption on $A$. $\endgroup$
    – YCor
    Jan 17 at 19:34
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    $\begingroup$ 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. $\endgroup$
    – YCor
    Jan 17 at 19:36
  • $\begingroup$ No assumption on $A$, just a finite-dimensional algebra over $k$. $\endgroup$ Jan 17 at 21:42

3 Answers 3


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.

  • $\begingroup$ 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. $\endgroup$ Feb 3 at 22:02
  • $\begingroup$ @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.] $\endgroup$
    – YCor
    Feb 4 at 16:01
  • $\begingroup$ This makes sense. Thanks for the clarification! $\endgroup$ 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.

  • $\begingroup$ 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"? $\endgroup$ Jan 18 at 10:07
  • $\begingroup$ @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 $k(n-k)$). $\endgroup$
    – 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 $A$.

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    $\begingroup$ 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 $\ker(f)=\ker(g)=\{0\}$. $\endgroup$
    – Yemon Choi
    Jan 18 at 1:46
  • $\begingroup$ I blundered. This assertion is valid given $f$ and $g$ are irreducible representations, but incorrect for any two homomorphisms. $\endgroup$
    – Onur Oktay
    Jan 18 at 2:30

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