TITLE: Are nearby subalgebras of matrix algebras conjugate?
QUESTION [7 upvotes]: 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 $A$.

REPLY [8 votes]: 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.