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

Question

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?

Answer 1

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.

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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.

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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).

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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: $f\begin{pmatrix}1 & \\ & \lambda\end{pmatrix} = f\begin{pmatrix}1 & \\ & |\lambda|\end{pmatrix}f\begin{pmatrix}1 & \\ & \frac \lambda {|\lambda|}\end{pmatrix}\in \Bbb R$.

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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").

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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 $\Bbb R^\times_+$.

Answer 2

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.

From this point on we return to the original argument..

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 $\alpha = 2 pi i$).

Answer 3

Note: We do have multiplicative from $M_n(\mathbb{H})$ to $\mathbb{H}$, since $M_n(\mathbb{H}) \subset M_{4n} (\mathbb{R})$, and compose with $\det$, however, the image is $\mathbb{R}$.

First, recall the formula for the product of two quaternions using the real and vector part:

$$(r_1 + v_1) \cdot (r_2 + v_2) = r_1 r_2 - \langle v_1, v_2 \rangle + (r_1 v_2 + r_2 v_2 + v_1 \times v_2)$$

It follows that if $q = r+ v \in \mathbb{H} \backslash \mathbb{R}$ ( i. e. $v\ne 0$), then the centralizer is a commutative $R$ subalgebra iso to $\mathbb{C}$:

$$C(q) = \mathbb{R} + \mathbb{R}v \simeq \mathbb{C}\subset \mathbb{H}$$

Say we have a morphism of groups from $GL(2, \mathbb{H})$ to $\mathbb{H}^{\times}$. Now, the elements $$\left(\begin{matrix} \alpha & 0 \\ 0 & 1\end{matrix} \right),\left(\begin{matrix} 1 & 0 \\ 0 & \beta\end{matrix} \right) $$ commute, for all $\alpha$, $\beta \in \mathbb{H}^{\times}$. We claim that this implies

$$\left (\begin{matrix}\alpha \gamma \alpha^{-1} \gamma^{-1} & 0 \\ 0 & 1 \end{matrix}\right)\mapsto 1 \in \mathbb{H}^{\times}$$

Indeed, if for some $\beta \in \mathbb{H}^{\times}$, the element $\left( \begin{matrix} 1 & 0 \\ 0 & \beta\end{matrix} \right ) $ does not map to $\mathbb{R}^{\times}$, then all $\left( \begin{matrix} \alpha & 0 \\ 0 & 1\end{matrix} \right ) $ map to a commutative subgroup. Otherwise, all $\left( \begin{matrix} 1 & 0 \\ 0 & \beta\end{matrix} \right ) $ map to $\mathbb{R}^{\times}$, the center of $\mathbb{H}^{\times}$, and so do the conjugate elements $\left( \begin{matrix} \alpha & 0 \\ 0 & 1\end{matrix} \right ) $.

Now let's see the conjugation :

$$\left( \begin{matrix} \delta & 0 \\ 0 & 1\end{matrix} \right ) \cdot \left( \begin{matrix} 1 & s \\ 0 & 1\end{matrix} \right )\cdot \left( \begin{matrix} \delta^{-1} & 0 \\ 0 & 1\end{matrix} \right ) =\left( \begin{matrix} 1 & \delta\cdot s \\ 0 & 1\end{matrix} \right ) $$

Now, let us recall that every quaternion $\delta$ of norm $1$ is a commutant: $$|\delta| = 1 \implies \delta = \alpha \gamma \alpha^{-1} \gamma^{-1} $$ in $\mathbb{H}^{\times}$. It follows from the above that if $|s_1| = |s_2|$ then $\left( \begin{matrix} 1 & s_1 \\ 0 & 1\end{matrix} \right )$, $\left( \begin{matrix} 1 & s_2 \\ 0 & 1\end{matrix} \right )$ map to the same element.

From here it follows that all $\left( \begin{matrix} 1 & s \\ 0 & 1\end{matrix} \right )$ map to $1$.

Now, using row reduction, one sees that every matrix in $GL(2, \mathbb{H})$ is of product of matrices of the form

$$\left(\begin{matrix} \alpha & 0 \\ 0 & 1\end{matrix} \right),\left(\begin{matrix} 1 & 0 \\ 0 & \beta\end{matrix} \right), \left( \begin{matrix} 1 & s \\ 0 & 1\end{matrix} \right ),\left( \begin{matrix} 1 & 0 \\ t & 1\end{matrix} \right ) $$

Provided that $|\alpha|= |\beta| = 1$, the above matrices map to $1\in \mathbb{H}^{\times}$. Since the diagonal matrices with $\alpha$, $\beta \in \mathbb{R}^{\times}$ commute, we conclude that the image of the morphism is commutative.