Are the rings $\mathbb{C}[X]/\langle X^2- c \mathrm{Tr}(X) X \rangle$ isomorphic when $c$ ranges over a neighborhood of 0?

Question

Let $X=(X_{IJ})_{I,J=1,\ldots,N}$ be a family of $N^2$ indeterminates and consider the ring $$ R_{N,c}=\mathbb{C}[X] / J_c,\quad J_c=\langle X^2 -c \mathrm{Tr}(X) X \rangle . $$ Here the notation means that one mods out the polynomial ring $\mathbb{C}[X]$ by the equalities implied by the corresponding equality of matrices. This is the coordinate ring (over $\mathbb{C}$) of the affine scheme of $N\times N$ matrices $M$ satisfying $M^2=c\mathrm{Tr}(M)M$.

In Bourget, Cabrera, Grimminger, Hanany, and Zhong - Brane Webs and Magnetic Quivers for SQCD, the authors made the claim that these rings $R_c$ are all isomorphic for all $|c|<1/N$. Is it true or not?

Answer 1

The isomorphism of the reduced underlying spaces is rather a trivial fact from undergraduate linear algebra. Let $X \in \mathcal{M}_{N}(\mathbb{C})$ such that $X^2 - \alpha \mathrm{Tr}(X).X = 0$, with $|\alpha| < \dfrac{1}{N}$. If $\mathrm{Tr}(X) = 0$ then $X$ is in the nilpotent cone, and vice-versa, if $X^2=0$, then $\mathrm{Tr}(X) = 0$.

Assume by absurd that $\mathrm{Tr}(X) \neq 0$, then $X$ is diagonalizable and its eigenvalues are $0$ and $\alpha \mathrm{Tr}(X)$. Let's denote by $m$ the multiplicity of $\alpha \mathrm{Tr}(X)$ as an eigenvalue of $X$, (note that $m \leq N$). Then we have:

$$ \mathrm{Tr}(X) = m \times \alpha \mathrm{Tr}(X),$$ which forces $\mathrm{Tr}(X) = 0$ since $|m \times \alpha|<1$, a contradiction. We thus find that for all $\alpha$ such that $|\alpha|< \dfrac{1}{N}$ the scheme of matrices defined by $X^2 - \alpha.\mathrm{Tr}(X).X = 0$ is isomorphic to the scheme of matrices defined by $X^2 = 0.$

EDIT : In a first version of my answer I claimed that my "topological" argument was sufficient to prove that the rings introduced by the OP are isomorphic. This not true as these rings are not reduced. Thus, an extra argument is needed to go from the isomorphism of the reduced underlying spaces to the isomorphisms of the actual schemes.

Answer 2

I don't think that these rings are isomorphic. If $\mathbb C[\epsilon]$ is the dual numbers then $3\times 3$ matrix of the form $$A=\begin{pmatrix} \lambda\epsilon & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & \mu \epsilon \end{pmatrix}$$ induces a map $R_c \rightarrow \mathbb C[\epsilon]$, for suitable choice of $\lambda, \mu \in \mathbb{C}$, but $A^2\neq 0$. This shows the ideals generated by $X^2$ and $X^2-c Tr(X) X$ are distinct, but does not exclude the existence of some other isomorphism.

Answer 3

I don't know if the answer to this question is still of interest, but I can prove that for $c$ in a $1$-dimensional (open) subscheme of $\mathbb{A}^1$, the projective schemes $\mathrm{Proj}(R_{N,c})$ are isomorphic. Unfortunately, the proof I have recquires some advanced techniques. I will still post it, as someone may see some drastic simplifcations that can be made.

First consider the linear map $\mathbb{C}[X] \longrightarrow \mathbb{C}[X]$ defined by: $$T_{\alpha} : M \longrightarrow T_{\alpha}(M) = M - \alpha \mathrm{Tr}(M).I_N.$$ It is a linear isomorphism provided $\alpha \neq \dfrac{1}{N}$. Furthermore, if one puts $M = T_{\alpha}(U)$, for $U = T_{\alpha}^{-1}(M)$, we have: $$ M^2 - c\mathrm{Tr}(M).M = 0 \ \Leftrightarrow \ U^2 - (2 \alpha + c(1-\alpha.N))\mathrm{Tr}(U).U + \alpha(c(1-\alpha.N)+\alpha) \mathrm{Tr}(U)^2.I_N=0$$ In particular, for $\alpha = \frac{c}{cN-1}$, we have: $$M^2 - c\mathrm{Tr}(M).M = 0 \ \Leftrightarrow U^2 - \frac{c}{c.N-1} \mathrm{Tr}(U).U = 0.$$ As $\alpha = \frac{c}{cN-1} \neq \dfrac{1}{N}$, the linear map $T_{\alpha}$ is an isomorphism and it maps isomorphically: $$T_{\alpha} : R_{N,\frac{c}{cN-1}} \stackrel{\sim}\longrightarrow R_{N,c}.$$

For $\alpha = \frac{1-s}{2}$, this transformation is the one exhibited by @Ycor in the case $N=2$.

We define the map $\varphi : \mathbb{C} \backslash \{\frac{1}{N} \} \longrightarrow \mathbb{C}$ by : $$\varphi(c) = \frac{c}{cN-1}$$ and we note that $\varphi \circ \varphi = \mathrm{id}$. The map $\varphi$ has exactly two fixed points : $0$ and $\frac{2}{N}$. If $c_0 \in \mathbb{C} \backslash \{0, \frac{2}{N} \}$, the sequence $(a_n = \varphi^{n}(c_0))_{n \geq 0}$ contains infintely many terms. The above argument then shows that the set:

$$\{c \in \mathbb{C}, \ \mathrm{Proj}(R_{N,c}) \simeq \mathrm{Proj}(R_{N,c_0}) \}$$ contains infinitely many points. I will now prove that this set is in fact a $1$-dimensional subscheme of $\mathbb{A}^1$. This is where I need some big guns, hopefully the argument may be simplified a lot.

Let $V \subset \mathrm{Proj}(\mathbb{C}[X]) \times \mathbb{C}$ the subscheme defined by the equation $X^2 - c\mathrm{Tr}(X)X = 0$ ($c$ is the name for the coordinate in the second factor) and let $p : V \longrightarrow \mathbb{C}$ the natural projection. By generic flatness, there exists a non-empty open subscheme $U \subset \mathbb{C}$, such that:

$$ p|_{p^{-1}(U)} : V_U = p^{-1}(U) \longrightarrow U$$ is flat. Note that $U$ is not equal to $\mathbb{C}$. Indeed $p^{-1}(0)$ is irreducible (but non-reduced!), while $p^{-1}(\frac{1}{N})$ contains (at least) two components of different dimension. We denote by $Z$ the complement of $U$ in $\mathbb{A}^1$, this is a closed (hence finite) subscheme of $\mathbb{C}$. As a consequence $U$ contains one of the $a_i$ defined above, say $a_{j_0}$.

We put $\mathcal{Q}$ be the Hilbert polynomial of $p^{-1}(a_{j_0}) = \mathrm{Proj}(R_{N,a_{j_0}}) \simeq \mathrm{Proj}(R_{N,c_0})$ and let $\mathcal{H}_{\mathcal{Q}}$ be the Hilbert scheme of subschemes of $\mathrm{Proj}(\mathbb{C}[X])$ having Hibert polynomial $\mathcal{Q}$. It is known that $\mathcal{H}_{\mathcal{Q}}$ is proper over $\mathbb{C}$, hence of finite type over $\mathbb{C}$.

The map $p : V_U \longrightarrow U$ being flat and the Hilbert polynomial of a fiber being $\mathcal{Q}$, the universal property of the Hibert scheme insures the existence of a map: $$ f : U \longrightarrow \mathcal{H}_{\mathcal{Q}}$$ such that the pull-back by $f$ of the universal family over $\mathcal{H}_{\mathcal{Q}}$ is $V_U$.

Now we consider the action of $\mathrm{G} = \mathrm{PGL}(\mathbb{C}^{n^2})$ on $\mathcal{H}_{\mathcal{Q}}$ and we let $\mathrm{G}.f(a_{j_0})$ be the orbit of $f(a_{j_0})$ under $\mathrm{G}$ in $\mathcal{H}_{\mathcal{Q}}$. This is a locally-closed finite type subscheme of $\mathcal{H}_{\mathcal{Q}}$ over $\mathbb{C}$.

The scheme $\mathrm{G}.f(a_{j_0})$ parametrizes all the subschemes of $\mathrm{Proj}(\mathbb{C}[X])$ which are linearly isomorphic to $f(a_{j_0})$. Furthermore, the property of $f$ with respect to the universal family of $\mathcal{H}_{\mathcal{Q}}$ and the construction of the sequence $(a_n)$ imply the inclusion: $$\{a_n, \ n \in \mathbb{N} \} \subset f^{-1}(\mathrm{G}.f(a_{j_0})).$$

As $\mathbb{A}^1$ is (obviously) affine and $\mathcal{H}_{\mathcal{Q}}$ is proper over $\mathbb{C}$, the scheme $f^{-1}(\mathrm{G}.f(a_{j_0}))$ is of finite type over $\mathbb{C}$. As a consequence $f^{-1}(\mathrm{G}.f(a_{j_0}))$ can not be zero-dimensional. Indeed, if it were, it would have infinitely many distincts connected components (the $\{a_n\}_{n \geq 0}$) and this is impossible for a scheme of finite-type over $\mathbb{C}$.

We deduce that $f^{-1}(\mathrm{G}.f(a_{j_0}))$ is positive dimensional in $\mathbb{A}^1$, and so it is Zariski dense. All fibers iof the form $p^{-1}(c)$, for $c \in f^{-1}(\mathrm{G}.f(a_{j_0})) \}$ are isomorphic, because $f^* W = V_U$ and all $W_{y}$, for $y \in \mathrm{G}.f(a_{j_0})\}$ are isomorphic, where: $$W \longrightarrow \mathcal{H}_{\mathcal{Q}}$$ is the universal family.

End of the proof.

(Minor) comment : it rather frustrating, but I can't prove that $0$ is the open subset where the $R_{N,c}$ are isomorphic to each other.