7
$\begingroup$

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 $A$.

$\endgroup$
6
  • 1
    $\begingroup$ Could you be explicit on what you mean by "this result"? $\endgroup$
    – YCor
    Jan 17, 2022 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, 2022 at 19:09
  • $\begingroup$ I understood this; what's missing is what's the assumption on $A$. $\endgroup$
    – YCor
    Jan 17, 2022 at 19:34
  • 3
    $\begingroup$ 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, 2022 at 19:36
  • $\begingroup$ No assumption on $A$, just a finite-dimensional algebra over $k$. $\endgroup$ Jan 17, 2022 at 21:42

3 Answers 3

11
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ 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, 2022 at 22:02
  • 1
    $\begingroup$ @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, 2022 at 16:01
  • $\begingroup$ This makes sense. Thanks for the clarification! $\endgroup$ Feb 4, 2022 at 21:53
8
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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, 2022 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, 2022 at 10:40
1
$\begingroup$

$\DeclareMathOperator\ker{ker}$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 $A$.

$\endgroup$
2
  • 1
    $\begingroup$ 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, 2022 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, 2022 at 2:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.