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Let $\mathbb R$ be the field of real numbers and $\mathbb C$ the field of complex numbers. It is well known that that $\mathbb C$ can be embedded in $M_2(\mathbb R)$.

This embedding can be extended in the obvious way to an embedding function $\varphi : M_n(\mathbb C)\rightarrow M_{2n}(\mathbb R)$.

My question is: consider the embedding $\varphi :M_2(\mathbb C)\rightarrow M_{4}(\mathbb R)$, it is possible to find an inner automorphism $\Psi : M_4(\mathbb R)\rightarrow M_{4}(\mathbb R)$ such that the intersection $$\varphi (M_2(\mathbb C))\cap \Psi(\varphi (M_2(\mathbb C)))$$

is the scalar matrices in $M_4(\mathbb R)$?

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Why specify that $\Psi$ is inner? By Skolem-Nother Theorem every automorphism of $M_4(\mathbb{R})$ is inner... –  Salvatore Siciliano Oct 9 '12 at 15:58
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1 Answer 1

This is an interesting problem. (According to V.I.Arnold, Misha Gromov said once: "There is only one way to find out whether a problem is good or not --- it simply must be solved!") First, let us reformulate it.

Given a finite-dimensional $\Bbb R$-linear space $V$ (in the original problem, $V$ is $4$-dimensional), denote by $M:=\text{Lin}_\Bbb R(V,V)$ the algebra of all $\Bbb R$-linear endomorphisms of $V$ and by $G:=\text{GL}V$, the group of $\Bbb R$-linear automorphisms of $V$. An $\Bbb R$-linear map $i\in M$ is said to be a complex structure on $V$ iff $i^2=-1$. An $\Bbb R$-linear map $h\in M$ is $\Bbb C$-linear with respect to a complex structure $i$ on $V$ iff $h$ commutes with $i$.

Problem. Is it possible to find $2$ complex structures $i_1,i_2$ on $V$ such that any $\Bbb C$-linear map with respect to both $i_1,i_2$ is just a multiplication by a real number? (Two complex structures on $V$ are conjugated by means of an $\Bbb R$-linear isomorphism because $\Bbb C$-linear spaces of the same dimension are isomorphic. Therefore, this problem is equivalent to the original one.) In other words, we are looking for $g\in G$ such that the centralizer of $i$ and $gig^{-1}$ in $M$ is trivial.

Remark. If such a $g$ exists, a generic $g$ also works. This means that, under the assumption of existing $g$ in question, there exists also $g\in M\setminus X$ with trivial centralizer of $i$ and $gig^{-1}$ for any proper Zariski closed $X\subset M$ (i.e., $M\ne X$ is given in $M$ by polynomial equations). Indeed, the condition on $g$ that the centralizer is nontrivial is Zariski closed (it is equivalent to vanishing some determinants) and a linear space over an infinite fieald cannot be a union of finitely many proper Zariski closed subsets (a well-known and easy fact).

Denote by $M_0$ the centralizer of $i$ in $M$ and let $a\in G$ be induced by the complex conjugation in a decomposition $V=\Bbb C\otimes_\Bbb RV_0$ compatible with $i$, $a^2=1$. Then $M=M_0\oplus aM_0$ is a $\Bbb Z/2$-grading, where $M_0=\{m\in M\mid imi^{-1}=m\}$ and $M_1:=aM_0=\{m\in M\mid imi^{-1}=-m\}$.

Answer. Suppose that $g=m+am_0\in G$ provides a trivial centralizer of $i$ and $gig^{-1}$ in $M$, where $m,m_0\in M_0$. By Remark, we can assume $m_0\in G$. As $m_0im_0^{-1}=i$, we can replace $g$ by $gm_0^{-1}$, i.e., we can assume $m_0=1$. By Remark, we can assume that $am+1\in G$ and that $(ma)^2$ does not belong to the centre of $M$. It follows from $a^2=1$ that $(am+1)^{-1}a(m+a)=1$. In other words, $g^{-1}=(am+1)^{-1}a$. Hence, $$gig^{-1}=i(m-a)(am+1)^{-1}a=$$ $$=i\big((m+a)(am+1)^{-1}a-2a(am+1)^{-1}a\big)=i\big(1-2(ma+1)^{-1}\big).$$ Therefore, the centralizer of $i$ and $gig^{-1}$ coincides with that of $i$ and $ma$. The latter contains $(ma)^2\in M_0$. A contradiction.

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