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It's well known the determinant provides a surjective homomorphism between $n \times n$ real matrices and $\mathbb{R}$ with the operation of multiplication.

I was curious of inter-matrix multiplicative homorphisms.

The trivial idea of viewing a $4\times 4$ matrix as a $2\times 2$ matrix with entries in the $2\times 2$ matrices suggests a determinant formula (of the usual type) but unfortunately that doesn't actually seem to work at all as a homomorphism after investigating it with sympy.

Doing some reading I stumbled upon: Quasideterminants for non-commutative rings.

But i'm a bit dissatisfied. I would like to think such a homomorphism should exist but I just don't know how to find it.

Sidharth Ghoshal
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    why not $A\mapsto \begin{bmatrix}\det A &0\0&\det A\end{bmatrix}$ – Shujian May 16 '24 at 02:31
  • This is a fair answer given what I asked. I added the surjective constraint to handle it. Thank you for pointing out that case :) – Sidharth Ghoshal May 16 '24 at 02:34
  • So ideas suggested here probably tell us that we might be able to classify ALL such maps by simply finding ONE such map, and then classifying all the non trivial automorphisms of $\mathbb{R}^{2\times 2}$. But to find ONE such map remains as a challenge. – Sidharth Ghoshal May 16 '24 at 02:46
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    Those answers implicitly assume that your map would preserve identities, which you seemingly don’t require. I would be shocked if such a function existed, but I have no reasoning for this. – Randall May 16 '24 at 03:40
  • Morphisms in what category? – Travis Willse May 16 '24 at 03:44
  • @TravisWillse Morphism here meaning homomorphism respecting multiplication. IE a homomorphism of monoids (since the matrices themselves form a monoid). I’ll update the title – Sidharth Ghoshal May 16 '24 at 03:45
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    @Randall If the map $\varphi: M_4(k) \to M_2(k)$ is required to be surjective and multiplicative, it automatically follows that $\varphi(I_4) = I_2$. Indeed, for all $A \in M_4(k)$, we have $\varphi(A) = \varphi(A)\varphi(I_4)$, and since $\varphi$ is surjective, the conclusion follows. – Haran May 16 '24 at 03:50
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    I see. So now we have that such a map must map invertibles to invertibles. – Randall May 16 '24 at 03:55
  • Maybe unhelpful comment: I would be surprised if this existed. Surjectivity here means every linear map $\mathbb{R}^2 \to \mathbb{R}^2$ comes from a linear map $\mathbb{R}^4 \to \mathbb{R}^4$. Ok not so crazy, we could do a silly projection maybe or something. But then we impose that the morphism must preserve composition! This seems very strict, but it's just intuition talking. I don't have a proof/example. Although maybe you could reframe and use that, i.e., matrices $\to$ elements of tensor product of dual spaces or something? – Derek Allums May 16 '24 at 04:36
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    This seems morally equivalent to asking whether $GL(2,\Bbb R)$ is a quotient of $GL(4,\Bbb R)$ (or the same with $SL$s instead of $GL$s). – Greg Martin May 16 '24 at 05:59

2 Answers2

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Claim: There exists no surjective multiplicative morphism from $M_r(k) \to M_s(k)$ for $r > s > 1$.

Proof: Assume the contrary, let $\varphi: M_r(k) \to M_s(k)$ be such a morphism. For all $A \in M_r(k)$, we have $\varphi(A) = \varphi(A)\varphi(I_r)$. The surjectivity of $\varphi$ allows us to pick $A$ such that $\varphi(A)$ is invertible. Thus, $\varphi(I_r) = I_s$. Furthermore, for all invertible $A \in M_r(k)$, we have $\varphi(A)\varphi(A^{-1}) = \varphi(I_r) = I_s$. Thus, $\varphi$ maps invertible transformations to invertible transformations.

Key Observation: For any $A, B \in M_r(k)$ with the same rank, there exist invertible $U, V \in M_r(k)$ satisfying $A = UBV$. Then, $\varphi(A) = \varphi(U)\varphi(B)\varphi(V)$. Since $\varphi(U)$ and $\varphi(V)$ are invertible, $\varphi(A)$ and $\varphi(B)$ have the same rank. In other words, the rank of $\varphi(A)$ is determined by the rank of $A$.

Let $N \in M_r(k)$ be nilpotent with rank $(r-1)$. The elements $I_r, N, \ldots, N^r$ have ranks $r, r-1, \ldots, 0$ respectively. Thus, the ranks of $I_s, \varphi(N), \ldots, \varphi(N)^r$ must include all possible integers from $0$ to $s$ by surjectivity. Furthermore, $\varphi(N)^r = 0_s$, so $\varphi(N)$ must be nilpotent. This forces $\varphi(N)$ to be nilpotent of rank $(s-1)$. We have consequently shown that for all $A \in M_r(k)$, $\mathrm{rank} \, \varphi(A) = \mathrm{max}(\mathrm{rank}\, A - (r-s), 0)$.

Now, let $e_1, \ldots, e_r$ be any basis for $k^r$. There are $t = {r \choose {r-s+1}} > s$ possible projections of rank $r - s + 1$ with respect to this basis. Note that we are using both $r > s$ and $s > 1$ for this to hold true. Denote these projections by $P_1, \ldots, P_t$. By our formula above, the images $Q_i = \varphi(P_i)$ have rank $1$ and they are projections by multiplicativity.

For each $1 \leqslant i \leqslant t$, pick an arbitrary eigenvector $v_i$ of eigenvalue $1$ for the projection $Q_i$ (so $v_i$ is determined up to scaling). Since $t > s$, the vectors $\{v_i\}$ must have some linear dependence. Assume without loss that $v_1$ is in the span of the rest. For each $i \neq 1$, since $P_1P_i = 0_r$, we have $Q_1Q_i = 0_s$. This implies that $Q_1Q_iv_i = Q_1v_i = 0$. However, $v_1$ is in the span of $\{v_i : i > 1\}$, so $Q_1v_1 = 0$. This yields the required contradiction!

Haran
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  • I've been digesting this and I'm just not seeing something. It's clear that if $N$ has rank $r-1$ then $\varphi(N)$ has some rank $<s$ since both $N$ and $\varphi(N)$ will be nilpotent. It's not clear at all that it will be even $s-1$ so your statement "must include all possible integers from 0 to $s$" does not follow for me. Do you know of a theorem which shows this MUST be the case? Additionally, is there any result which states if $A$ is nilpotent of rank $k$ then $A^{r}$ is nilpotent of rank $k-r$? These probably exist as undergrad theorems but I can't seem to find them. – Sidharth Ghoshal May 18 '24 at 23:08
  • We prove that the rank of $\varphi(A)$ only depends on the rank of $A$. Since the powers of $N$ have all possible ranks and $\varphi$ is surjective, the powers of $\varphi(N)$ must have all possible ranks as well. Moreover, $\varphi(N)$ is nilpotent, so the rank of its powers strictly decrease until reaching zero. This gives the required claim. – Haran May 18 '24 at 23:11
  • It is not true that if $A$ has rank $k$, then $A^r$ has rank $k-r$. The way in which the rank of the powers of some nilpotent operator decrease depends on its Jordan block decomposition. In the specific case, where you have a nilpotent operator which has rank one less than full rank, there must precisely be one Jordan block, so the rank will decrease by $1$ until it reaches zero. You might want to try a few examples out! – Haran May 18 '24 at 23:13
  • oh i see. So if some Rank $r-1$ matrix is mapped to a $s-2$ matrix then all $r-1$ rank matrices must be mapped as such. Okay so that is likely not possible. Suppose that is true for the sake of contradiction. We want to argue that the surjective condition was broken. But it's still not clear to me that something weird could happen, for example rank $r-1$ matrices could map to rank $s-2$ yet rank $r-2$ matrices could map rank $s-1$. We need a $\text{Rank}(A) \le \text{Rank}(B) \rightarrow \text{Rank}(\varphi(A)) \le \text{Rank}(\varphi(B)) $ – Sidharth Ghoshal May 18 '24 at 23:19
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    So we also have a rank $r-2$ matrix in hand, namely $N^2$. Since $\varphi(N^2) = \varphi(N)^2$ certainly has rank lower than $\varphi(N)$, our claim holds true. – Haran May 18 '24 at 23:20
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    Ah this is actually really slick! the ability to use any rank $k$ matrix to describe them all of them pulls this together. :) – Sidharth Ghoshal May 18 '24 at 23:20
  • What are these projections exactly? There are $\begin{pmatrix} r \ s \end{pmatrix} = \begin{pmatrix} r \ r-s \end{pmatrix} $ projections from $k^r$ to $s^r$. OTOH there are are $2^r$ projections in general from $k^r$ to some subset of the basis. Neither of these is $\begin{pmatrix} r \ r - s + 1 \end{pmatrix} $ so I don't think I know what your target space for projection is. – Sidharth Ghoshal May 18 '24 at 23:28
  • Let us continue this discussion in chat. – Haran May 18 '24 at 23:29
  • sure! just joined :) – Sidharth Ghoshal May 18 '24 at 23:29
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If there is a surjective multiplicative homomorphism $\varphi$ from $M_r(k)$ to $M_s(k)$ where $r>s>1$, then

  • $\varphi(I_r)$ is a multiplicative identity on the range of $\varphi$, hence $\varphi(I_r)=I_s$, and further $\varphi(GL_r(k))\subseteq GL_s(k)$.

  • (Simliar to what @Haran did) If $A\in M_r(k)$ is not invertible, then there exists invertible $U, V$ such that $B=UAV$ is nilpotent, therefore $\varphi(UAV)=\varphi(U)\varphi(A)\varphi(V)$ is also nilpotent, and since $\varphi(U), \varphi(V)$ are invertible, $\varphi(A)$ is not. Hence $\varphi(GL_r(k))=GL_s(k)$.

But there is no such surjective group homomorphism from $GL_r(k)$ to $GL_s(k)$.

Just a user
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