There is no continuous surjective multiplicative map from $M_n(\mathbb H)$ to $\mathbb H$

Let $$\mathbb H$$ denote the field of quaternions. I would like to prove that there does not exist any function $$f:M_n(\mathbb H)\rightarrow \mathbb H$$ for $$n\geq 2$$ that is continous surjective and multiplicative.

I have been thinking about this problem for a while but I can't find any contradiction assuming that such a function does exist. I tried considering preimages for $$1,i,j,k$$ and toying with them, I also tried infering the values of some specific matrices (the $$\lambda I_n$$, the nilpotent matrices, etc...) but I couldn't reach any conclusion. Mostly, I fail to see how to make use of the continuity here.

Would somebody have a hint as to how to proceed with this problem?

• How is the situation different from the case where we replace $\mathbb{H}$ with $\mathbb{R}$ or $\mathbb{C}$? Dec 13 '19 at 9:38
• Do you know about representations of Lie groups? If so, restrict such a map to get a representation of $GL_n(\mathbb H)$ over $\mathbb H$. Tensoring over $\mathbb C$ gives a 2-dimensional complex representation of $\GL_{2n}(\mathbb C)$. This must be reducible if $n > 1$, which contradicts surjectivity. Dec 14 '19 at 20:49
• There used to be a very good answer, due to ThorstenK to the question in my second comment. Somehow it disappeared so I will reproduce it here. There is a multiplicative non-zero map by composing the embedding of $Mat(n, \mathbb{H})$ into $Mat(4n, \mathbb{R})$ with the determinant and the embedding of $\mathbb{R}$ into $\mathbb{H}$. Dec 15 '19 at 18:43
• It would be nice to know the subgroup structure of $GL(n, \mathbb{H})$. By dimension considerations the kernel of the map $f: GL(n, \mathbb{H}) \to GL(1, \mathbb{H}) \cong SU(2) \times \mathbb{R}_{>0}$ must be pretty big and when the only pretty big subgroup of $GL(n, \mathbb{H})$ turns out $GL(n, \mathbb{H})$ itself we get a contradiction with surjectivity. However I could not find much information on $GL(n, \mathbb{H})$ as a group on the internet and the somewhat similar group $GL(n, \mathbb{C})$ does have a fairly big subgroup in the form of $SL(n, \mathbb{C})$ so that doesn't bode well... Dec 16 '19 at 20:07
• @Vincent Thank you Vincent for these explanations, I understand the argument now. Dec 17 '19 at 0:10

I will work out the case $$n = 2$$ in detail. The same proof works for general $$n$$, I just want to save the labor of typing $$n$$ by $$n$$ matrices...

Thus assume that $$f:\operatorname M_2 = \operatorname M_2(\Bbb H) \rightarrow \Bbb H$$ is a surjective multiplicative map.

Lemma 1. Whenever $$A\in \operatorname M_2$$ is invertible, the image $$f(A)$$ is also invertible.

Proof: If $$A$$ is invertible, then multiplication by $$A$$ is a bijection on $$\operatorname M_2$$. Hence $$f(A)$$ cannot be zero, otherwise $$f$$ is constantly zero.

Lemma 2. The map $$f$$ restricted to $$\operatorname{GL}_2 = \operatorname{GL}_2(\Bbb H)$$ gives a group homomorphism from $$\operatorname{GL}_2$$ to $$\Bbb H^\times$$. In particular, we have $$f\begin{pmatrix} 1 & \\ & 1\end{pmatrix} = 1$$.

Proof: This is clear from Lemma 1.

We want to arrive at a contradiction, hence showing that such an $$f$$ does not exist.

Assumption 3. Without loss of generality, we may assume that $$f\begin{pmatrix} & 1\\1 &\end{pmatrix} = 1$$.

Note: for general $$n$$, we have the canonical embedding of the symmetric group $$S_n$$ into $$\operatorname{GL}_n$$, and this assumption becomes: $$f(\sigma) = 1$$ for all $$\sigma \in S_n$$.

Why we can make this assumption: we have $$f(\sigma)^{n!} = f(\sigma^{n!}) = 1$$ by Lemma 2, hence by changing $$f$$ to $$f^{n!}$$, which is still surjective multiplicative, we may make this assumption.

From now on, we always make Assumption 3.

Lemma 4. We have $$f\begin{pmatrix}1 & \\ & \lambda\end{pmatrix} = f\begin{pmatrix}\lambda & \\ & 1\end{pmatrix}$$ for any $$\lambda \in \Bbb H$$.

Proof: This comes from the identity $$\begin{pmatrix}1 & \\ & \lambda\end{pmatrix}\begin{pmatrix} & 1\\1 & \end{pmatrix} = \begin{pmatrix} & 1\\1 & \end{pmatrix}\begin{pmatrix}\lambda & \\ & 1\end{pmatrix}$$ and Assumption 3.

Lemma 5. We have $$f\begin{pmatrix}1 & \\ & z\end{pmatrix} = 1$$ for all $$z \in \Bbb H^\times$$ with $$|z| = 1$$.

Proof: For any $$\lambda, \mu \in \Bbb H^\times$$, we have: $$f\begin{pmatrix}1 & \\ & \lambda\mu\end{pmatrix} = f\begin{pmatrix}1 & \\ & \lambda\end{pmatrix}f\begin{pmatrix}1 & \\ & \mu\end{pmatrix}=f\begin{pmatrix}\lambda & \\ & 1\end{pmatrix}f\begin{pmatrix}1 & \\ & \mu\end{pmatrix}=f\begin{pmatrix}\lambda & \\ & \mu\end{pmatrix} = f\begin{pmatrix}1 & \\ & \mu\end{pmatrix}f\begin{pmatrix}\lambda & \\ & 1\end{pmatrix} = f\begin{pmatrix}1 & \\ & \mu\end{pmatrix}f\begin{pmatrix}1 & \\ & \lambda\end{pmatrix} = f\begin{pmatrix}1 & \\ & \mu\lambda\end{pmatrix}.$$ Therefore we have $$f\begin{pmatrix}1 & \\ & \lambda\mu\lambda^{-1}\mu^{-1}\end{pmatrix} = 1$$. But any $$z \in \Bbb H^\times$$ with $$|z| = 1$$ can be written as $$\lambda\mu\lambda^{-1}\mu^{-1}$$ for some $$\lambda, \mu \in \Bbb H^\times$$.

Lemma 6. For any $$a\in \Bbb R$$, the value $$f\begin{pmatrix}a & \\ & a\end{pmatrix}$$ is real.

Proof: Since the matrix $$A = \begin{pmatrix}a & \\ & a\end{pmatrix}$$ is in the center of $$\operatorname M_2$$, we have $$f(A)f(B) = f(AB) = f(BA) = f(B)f(A)$$ for all $$B\in \operatorname M_2$$. The surjectivity of $$f$$ then implies that $$f(A)$$ lies in the center of $$\Bbb H$$, namely $$\Bbb R$$.

Assumption 7. Without loss of generality, we may assume that $$f\begin{pmatrix}1 & \\ & a\end{pmatrix}$$ is real for all $$a \in \Bbb R$$.

Why we can make this assumption: we already have $$\left(f\begin{pmatrix}1 & \\ & a\end{pmatrix}\right)^2 = f\begin{pmatrix}1 & \\ & a\end{pmatrix}f\begin{pmatrix}a & \\ & 1\end{pmatrix} = f\begin{pmatrix}a & \\ & a\end{pmatrix}\in \Bbb R.$$Therefore, by changing $$f$$ to $$f^2$$, we may make this assumption (while still keeping all required properties of $$f$$, including Assumption 3).

From now on, we always make Assumptions 7.

Lemma 8. We have $$f\begin{pmatrix}1 & \\ & \lambda\end{pmatrix}\in \Bbb R$$ for all $$\lambda \in \Bbb H$$.

Proof: The case $$\lambda = 0$$ is covered by Assumption 7. For $$\lambda \neq 0$$, by Lemma 5 and Assumption 7, we have: $$f\begin{pmatrix}1 & \\ & \lambda\end{pmatrix} = f\begin{pmatrix}1 & \\ & |\lambda|\end{pmatrix}f\begin{pmatrix}1 & \\ & \frac \lambda {|\lambda|}\end{pmatrix}\in \Bbb R$$.

Lemma 9. For any $$\alpha \in \Bbb H^\times$$, we have $$f\begin{pmatrix}1 & \alpha \\ & 1\end{pmatrix} = f\begin{pmatrix}1 & 1\\ & 1\end{pmatrix}$$.

Proof: This comes from the identity $$\begin{pmatrix}1 & \\ & \alpha^{-1}\end{pmatrix}\begin{pmatrix}1 & 1\\ & 1\end{pmatrix}\begin{pmatrix}1 & \\ & \alpha\end{pmatrix} = \begin{pmatrix}1 & \alpha\\ & 1\end{pmatrix}$$ and the fact that $$f\begin{pmatrix}1 & \\ & \alpha\end{pmatrix}$$ is a real number, hence is in the center of $$\Bbb H$$.

Lemma 10. For any $$\alpha \in \Bbb H$$, we have $$f\begin{pmatrix}1 & \alpha \\ & 1\end{pmatrix} = 1$$.

Proof: Let $$h$$ be the value of $$f\begin{pmatrix}1 & 1 \\ & 1\end{pmatrix}$$. By Lemma 9, we have $$h = f\begin{pmatrix}1 & 2 \\ & 1\end{pmatrix} = h^2$$. By Lemma 1, we get $$h = 1$$ and Lemma 9 tells us that $$f\begin{pmatrix}1 & \alpha \\ & 1\end{pmatrix} = 1$$ for any $$\alpha \in \Bbb H^\times$$. The case $$\alpha = 0$$ is Lemma 2.

Conclusion. The map $$f$$ takes values in $$\Bbb R$$ on $$\operatorname M_2$$, hence is not surjective. We obtain a contradiction.

Proof: Just note that any matrix in $$\operatorname M_2$$ can be written as a product of matrices of the form $$\begin{pmatrix}1 & \alpha \\ & 1\end{pmatrix}$$, $$\begin{pmatrix} & 1 \\1 & \end{pmatrix}$$, $$\begin{pmatrix}1 & \\ & \lambda\end{pmatrix}$$ with $$\alpha, \lambda \in \Bbb H$$ (by performing "row and column operations").

Final remarks.

• As claimed in the very beginning, the proof adapts without difficulty to general $$n$$.

• The continuous assumption is not used. All arguments are algebraic.

• Since it's a proof by contradiction, it doesn't show that any multiplicative map from $$\operatorname M_n(\Bbb H)$$ to $$\Bbb H$$ has image in $$\Bbb R$$. But it is true that any group homomorphism from $$\operatorname{GL}_n(\Bbb H)$$ to $$\Bbb C^\times$$ must factorize through $$\Bbb R^\times_+$$, as the abelianization of $$\operatorname{GL}_n(\Bbb H)$$ is isomorphic to $$\Bbb R^\times_+$$.

• Very nice proof, great use of proof by contradiction. I only do not agree with the last final remark: the map that you describe does give non-negative real entries, as described here: math.stackexchange.com/q/3477987/101420. It would be nice to see if an example exists of a non-surjective multiplicative map with actual non-real elements in the image. Dec 20 '19 at 21:14
• @Vincent After thinking more about it, I think there is no example of group homomorphism from $\operatorname{GL}_n(\Bbb H)$ to $\Bbb C^\times$ taking values outside $\Bbb R$. This is because the abelianization of $\operatorname{GL}_n(\Bbb H)$ is isomorphic to $\Bbb R^\times_{> 0}$. More precisely, the homomorphism $\Bbb H^\times \rightarrow \operatorname {GL}_n(\Bbb H)$ sending $z$ to $diag(z, 1, \dotsc, 1)$ induces an isomorphism between the abelianizations. This is not hard to prove, and is in fact known to Dieudonné: it's Theorem 1 here: numdam.org/article/BSMF_1943__71__27_0.pdf Dec 20 '19 at 23:02
• Hmm I just concluded that image outside $\mathbb{R}$ is possible, see the last remark in my answer. Put more simply: take a homomorphism to $\mathbb{R}_+$, take the log to make it an homomorphism to the additive group $\mathbb{R}$ and wind that around the unit circle in $\mathbb{C}$... Dec 20 '19 at 23:41
• @Vincent Ah, of course, there are non-trivial homomorphisms from $\Bbb R^\times_+$ to $\Bbb C^\times$... It just has to factor through $\Bbb R^\times_+$. Dec 20 '19 at 23:58
• Thanks for the acceptance. I believe that we all enjoyed solving and discussing this problem. Dec 21 '19 at 23:18

This is a modified version of my earlier answer with some gaps filled up, hence the weird numbering a Let $$D \subset Mat(2, \mathbb{H})$$ denote the group of invertible diagonal matrices. Let $$L$$ be the group of lower triangular matrices with $$1$$s on the diagonal and let $$U$$ be the group of upper triangular matrices with $$1$$s on the diagonal. Let $$G \subset GL(2, \mathbb{H})$$ the set of of matrices $$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ such that $$a \neq 0$$ an $$d - ca^{-1}b \neq 0$$. The set $$G$$ is open and dense in $$GL(2, \mathbb{H})$$ so that once we understand the image $$f(G)$$ of $$G$$ under a continuous multiplicative map, we also understand the image of $$GL(2, \mathbb{H})$$. Alternatively if you don't like topology you can show that every matrix in $$GL(2, \mathbb{H})$$ can be written as a product of matrices from $$G$$. The reason for working with $$G$$ is that every $$g \in G$$ can be decomposed as

$$g = ldu$$ for some $$l \in L, d \in D, u \in U \qquad (1)$$

Let's focus on $$D$$ first. It has two subgroups $$D_1$$ and $$D_2$$ consisting respectively of the diagonal matrices with a 1 in the lower right corner and those with a 1 in the upper left corner. As a group both $$D_1$$ and $$D_2$$ are isomorphic to $$\mathbb{H}^*$$ of course. We draw some conclusions from the group structure of $$\mathbb{H}^*$$.

Lemma 0: The group $$\mathbb{H}^*$$ decomposes as a direct product of topological groups $$\mathbb{H}^* \cong SU(2) \times \mathbb{R}_+$$ where the cannonical projection onto the second term is just the familiar modulus operator $$|.|$$ and the subgroup $$SU(2)$$ appears as the set of elements of norm $$1$$.

Lemma 0.5 The group $$SU(2)$$ is almost simple: its only normal subgroups are $$\{1\}, \{-1, +1\}$$, and $$SU(2)$$ itself. The quotient $$SU(2)/\{-1, 1\}$$ is isomorphic to $$SO(3)$$ which is simple and is not isomorphic to any subgroup of $$\mathbb{H}$$.

Corollary 0.75: every group homomorphism from $$\mathbb{H}^*$$ to itself maps the $$SU(2)$$-subgroup of norm 1 elements in the domain either bijectively onto the $$SU(2)$$-subgroup of norm 1 elements in the codomain or onto the one element subgroup $$\{1\}$$ in the codomain.

Lemma 1, modified: let $$f$$ be a multiplicative map from $$D \to \mathbb{H}^*$$. Then for at least one of the two subgroups $$D_1, D_2$$ it maps the $$SU(2)$$-subgroup of norm 1 elements inside that subgroup to $$\{1\}$$.

Proof: Let $$y_1, x_y$$ be two non-commuting elements of $$SU(2)$$ in the codomain $$\mathbb{H}$$. If $$f$$ does not map the norm 1 elements in $$D_1$$ to $$1$$ then, by corollary 0.75 there is an $$x_1 \in D_1$$ with $$f(x_1) = y_2$$. Similarly if $$f$$ does not map the norm 1 elements in $$D_2$$ to $$1$$ then there is a $$x_2$$ with $$f(x_2) = y_2$$. Now $$x_1x_2 = x_2x_1$$ since every element in $$D_1$$ commutes with every element in $$D_2$$ but $$f(x_1)f(x_2) \neq f(x_2)f(x_1)$$, a contradiction.

The question is now what happens to the $$\mathbb{R}_+$$ subgroup of that group ($$D_1$$ or $$D_2$$). I thought that it must be mapped to $$\mathbb{R}_+$$ in $$\mathbb{H}^+$$ but that is incorrect, $$D_1 \cong \mathbb{H}^*$$ can be mapped into a spiral via e.g. $$f(x) = e^{(a + bi)\log(|x|)}$$ while still sending $$SU(2)$$ to $$\{1\}$$, the latter condition being equivalent to $$f(x) = f(y)$$ whenever $$|x| = |y|$$ as in lemma 0.

However what we do know is that if the restriction of $$f$$ maps the $$SU(2)$$ part of $$D_i$$ (for some $$i \in \{1, 2\}$$) to $$1$$ and hence only depends on its restriction to the $$\mathbb{R}_+$$ part then $$f(x)f(y) = f(y)f(x)$$ for every $$x,y \in D_i$$. It follows that $$f(D_i)$$ is contained in a two dimensional subalgebra $$\mathbb{C}'$$ of $$\mathbb{H}$$ isomorphic to $$\mathbb{C}$$.

Now let $$f: GL(2, \mathbb{H}) \to \mathbb{H}^*$$ be a multiplicative map and let $$J = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$. Since $$J^2 = 1$$ we have that either $$f(J) = 1$$ or $$f(J) = -1$$. But since $$JD_1J = D_2$$ and vice versa we conclude from Lemma 1 and the above subsequent reasoning that:

Lemma 2, modified: every multiplicative map $$f: GL(2, \mathbb{H}) \to \mathbb{H}^*$$ maps $$D$$ into $$\mathbb{C}' \backslash \{0\}$$ for some 2-dimensional subalgebra $$\mathbb{C}' \subset \mathbb{H}$$. Moreover $$f(D) = f(D_1)$$ hence if $$f$$ is continuous the image $$f(D)$$ is connected and at most one dimensional.

Progress! Before moving on to $$L$$ and $$U$$ we collect some corollaries of this result.

Corollary 2.5: Let $$f: GL(2, \mathbb{H}) \to \mathbb{H}^*$$ be a multiplicative map and $$x = \begin{pmatrix} a & 0 \\ 0 & d \end{pmatrix} \in D$$. Then $$f(x) = f(|x|)$$ where $$|x| \in D$$ is the matrix whose entries are the absolute values of the entries in $$x$$.

Proof: by Lemma 2 the space $$f(D)$$ is too small to contain $$SU(2)$$, so both the copy of $$SU(2)$$ inside $$D_1$$ and that in $$D_2$$ are mapped to 1. The result then follows from Lemma 0.

Corollary 3 (same result, new proof): Let $$f: GL(2, \mathbb{H}) \to \mathbb{H}^*$$ be a multiplicative map and $$q \in D$$. Then $$f(qxq^{-1}) = f(x)$$ for all $$x \in GL(2,\mathbb{H})$$.

Proof: we distinguish two cases. Either $$f(D) \subset \mathbb{R}$$ or it doesn't. In the first case we have that $$f(q)$$ commutes with $$f(x)$$ for every $$x \in \ GL(2, \mathbb{H})$$ and we have $$f(qxq^{-1} = f(q)f(x)f(q^{-1} = f(x)f(q)f(q^{-1}) = f(x)$$. In the second case we have a $$y \in D$$ such that $$f(y) \not\in \mathbb{R}$$. Let $$|y|$$ be as in the previous corollary and let $$r \in D$$ be the matrix whose entries are the square roots of the corresponding entries of $$|y|$$. We see from the previous corollary that $$f(r)^2 = f(y)$$. Let $$s = r(JrJ)$$. Then $$s$$ is a real scalar multiple of the identity matrix but $$f(s) = f(r)^2 = f(y) \not\in \mathbb{R}$$. Since $$s$$ is a real scalar multiple of the identity matrix we have that $$sx = xs$$ and hence $$f(s)f(x) = f(x)f(s) \qquad(1.5)$$ for every $$x \in GL(2, \mathbb{H})$$. But since $$f(s)$$ is a non-real element of $$\mathbb{C}'$$, with $$\mathbb{C}'$$ as in Lemma 2 we find that (1.5) implies that $$f(x) \in \mathbb{C}'$$ for every $$x \in GL(2, \mathbb{H})$$. It then follows from lemma 2 that $$f(q)f(x) = f(x)f(q)$$ for every $$q \in D$$ and the claim of Corollary 3 follows.

We use corollary 3 to understand the action of $$f$$ on $$U$$.

Lemma 4: let $$f: GL(2, \mathbb{H}) \to \mathbb{H}^*$$ be a multiplicative map and let $$u_1, u_2 \in U \backslash \{I\}$$. Then $$f(u_1) = f(u_2)$$.

Proof: $$u_i = \begin{pmatrix} 1 & b_i \\ 0 & 1 \end{pmatrix}$$ for $$i = 1, 2$$ where $$b_1, b_2$$ are non-zero, hence invertible, elements of $$\mathbb{H}$$.

In general we have $$\begin{pmatrix} a & 0 \\ 0 & d \end{pmatrix} \begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix}\begin{pmatrix} a^{-1} & 0 \\ 0 & d^{-1} \end{pmatrix} = \begin{pmatrix} 1 & abd^{-1} \\ 0 & 1 \end{pmatrix} \qquad (2)$$

Taking $$a = b_2, b = d = b_1$$ in (2) we obtain $$qu_1q^{-1} = u_2$$ where $$q \in D$$ is the diagonal matrix $$\begin{pmatrix} a & 0 \\ 0 & d \end{pmatrix}$$ from (2). The claim $$f(u_1) = f(u_2)$$ then follows from Corollary 3.

Corollary 5: let $$f: GL(2, \mathbb{H}) \to \mathbb{H}^*$$ be a continuous multiplicative map, then $$f(U) = \{1\}$$.

Proof: by Lemma 4 we have that $$f$$ takes only one value on $$U \backslash \{I\}$$ and hence by continuity it should take this same value on $$I \in U$$ as well. But this means we know the unique value that $$f$$ takes on $$U$$ because $$f(I) = 1$$ for any multiplicative map.

In a completely analogous way we get:

Lemma 6: let $$f: GL(2, \mathbb{H}) \to \mathbb{H}^*$$ be a continuous multiplicative map, then $$f(L) = \{1\}$$.

Now we can prove our main result.

Theorem 7, modified: let $$f: Mat(2, \mathbb{H}) \to \mathbb{H}$$ be a continuous multiplicative map, then either $$f$$ maps every element of $$Mat(2, \mathbb{H})$$ to zero, or it maps invertible matrices to a one dimensional multiplicative Lie subgroup of $$\mathbb{H}^*$$ contained in a two dimensional subalgebra $$\mathbb{C}'$$ of $$\mathbb{H}$$ isomorphic to $$\mathbb{C}$$.

Proof: We distinguish two cases: either $$f(GL(2, \mathbb{H})) \subset \mathbb{H}^*$$ or there is some $$g \in GL(2, \mathbb{H})$$ with $$f(g) = 0$$. In the latter case we find that $$f$$ is identically zero as $$f(x) = f(xg^{-1}g) = f(xg^{-1})f(g) = f(xg^{-1})0 = 0$$ for every $$x \in Mat(2, \mathbb{H})$$. In the second case, let $$g \in GL(2, \mathbb{H})$$. As in the text preceding (1) we may assume that $$g \in G$$, with $$G$$ defined there. From (1) and Corollary 5 and Lemma 6 we see that there is a $$d \in D$$ such that $$f(g) = f(d)$$. Lemma 2 then gives us the claim of the theorem.

I like Theorem 7 because it tells us that yes, non-zero maps may exist, but only under very severe restrictions. To get the full result we only need:

Lemma 8: The set $$GL(2, \mathbb{H})$$ of invertible matrices is dense (in the topological sense) in the real vectorspace $$Mat(2, \mathbb{H})$$ of all matrices.

Remark 9: I think that every group homomorphism $$f: \mathbb{R}_+ \to \mathbb{H}^*$$ is of the form $$x \mapsto \exp(\alpha \log x)$$ for some quaternion $$\alpha$$. (Here $$\exp$$ is defined by the same power series as always.) Reading my proof with this in mind we find that for non-zero continous multiplicative $$f$$ we find that there is an $$\alpha \in \mathbb{H}$$ such that for each $$g = \begin{pmatrix} a & b \\ c& d \end{pmatrix} \in G$$ we have that $$f(g) = \exp(\alpha \log(|a||d - ca^{-1}b|))$$. Now we can recognize the expression inside the $$\log$$ as the determinant of the $$(2 \times 2) \times (2 \times 2)$$-block matrix $$g'$$ over $$\mathbb{C}$$ associated to $$g$$ in the standard way (i.e. as in your linked question). By continuity we then conclude that $$f(g) = \exp(\alpha \log(\det(g')))$$ for every $$g \in Mat(2, \mathbb{H})$$. This then gives a nice classification of all possible $$f$$ and answers the question about the existence of $$f$$ with non-real image (e.g. take $$\alpha = 2 pi i$$).

• Thank you for your proof Vincent. I noticed no flaw in it, so I believe it to be true. You proved a stronger statement than the problem I initially asked in the case of $n=2$. Using your theorem $7$ and if Gauß reduction works over quaternionic matrices (which I am not exactly sure), we can actually drop the continous hypohesis in my initial problem : any multiplicative map from $M_n(\mathbb H)$ (for now, $n=2$) to $\mathbb H$ is not surjective. Indeed, with Gauß reduction we could argue that any matrix is equivalent to $J_r$ (the diagonal matrix with $r$ times the entry $1$ then all $0$)[...] Dec 20 '19 at 5:55
• [...] and then, by multiplicativity and your theorem $7$, we see that all such maps could only take values inside the union of at most $n$ half-lines inside $\mathbb H$. Each half lines would be directed by the image of $J_r$, for $r$ varying from $1$ to $n$. This would be a really nice result to me ; and it could work for general $n$ provided that we can extend your theorem $7$ and that we can apply Gauß reuduction on quaternionic matrices. Dec 20 '19 at 5:57
• Unfortunately I noticed that there is a flaw, in Lemma 1, although I believe with some extra work we can get back on track from Corollaray 3 onwards. I'll try and edit the answer tonight (European time) Dec 20 '19 at 9:36