343: n-Dimensional Quaternion General Linear Group Is 'Groups - Homomorphism Morphisms' Isomorphic to Set of Nonzero Determinant Corresponding 2n x 2n Complex Matrices and Can Be Represented by Latter

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A description/proof of that n-dimensional quaternion general linear group is 'groups - homomorphism morphisms' isomorphic to set of nonzero determinant corresponding 2n x 2n complex matrices and can be represented by latter

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Topics

About: group

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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Starting Context

• The reader knows a definition of quaternion.
• The reader knows a definition of general linear group.
• The reader admits the proposition that the set of the quaternions is 'rings - homomorphism morphisms' isomorphic to the set of the corresponding 2 x 2 complex matrices.
• The reader admits the proposition that the set of the n x n quaternion matrices is 'rings - homomorphism morphisms' isomorphic to the set of the corresponding 2n x 2n complex matrices, via the between-the-set-of-the-quaternions-and-the-set-of-the-corresponding-2-x-2-complex-matrices 'rings' isomorphism.
• The reader admits the proposition that the multiplication of any matrix made of any same size blocks by any matrix made of blocks of any multiplicable (with blocks of the former matrix) same size is blocks-wise.

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Target Context

• The reader will have a description and a proof of the proposition that the n-dimensional quaternion general linear group is 'groups - homomorphism morphisms' isomorphic to the set of the nonzero determinant corresponding 2n x 2n complex matrices, and can be represented by the latter, which means that the map of each element of the former is represented by the map of the corresponding element of the latter.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Description

1st, let us define some symbols. $\mathbb{H}$ is the set of the quaternions. $GL(n,\mathbb{H})$ is the n-dimensional quaternion general linear group. $V$ is the n-dimensional vectors space over $\mathbb{H}$ on which $GL(n,\mathbb{H})$ operates. $M_n(\mathbb{H})$ is the set of the n x n quaternion matrices. $(M_n(\mathbb{H}){)}^x$ is the set of the invertible n x n quaternion matrices. $f$ is the 'rings - homomorphism morphisms' isomorphism, $f:M_n(\mathbb{H})\to f^{″}(M_n(\mathbb{H}))$, defined in the proposition that the set of the n x n quaternion matrices is 'rings - homomorphism morphisms' isomorphic to the set of the corresponding 2n x 2n complex matrices, via the between-the-set-of-the-quaternions-and-the-set-of-the-corresponding-2-x-2-complex-matrices 'rings - homomorphism morphisms' isomorphism, where $f^{″}$ is defined there. $(f^{″}(M_n(\mathbb{H})){)}^x$ is the set of the nonzero determinant matrices in $f^{″}(M_n(\mathbb{H}))$. $h^{\prime }$ is the 'vectors spaces - linear morphisms' isomorphism, $h^{\prime }:V\to {\mathbb{H}}^n$, when a basis for $V$ is chosen. $h$ is the 'groups - homomorphism morphisms' isomorphism, $GL(n,\mathbb{H})\to (M_n(\mathbb{H}){)}^x$, corresponding to the chosen basis. $f^{‴}$ is the map from ${\mathbb{H}}^n$ to $f^{‴}({\mathbb{H}}^n)\subseteq {\mathbb{C}}^{2nx2}$ that replaces each component of each ${\mathbb{H}}^n$ element by $f^{\prime }$, which is the between-the-set-of-the-quaternions-and-the-set-of-the-corresponding-2-x-2-complex-matrices 'rings - homomorphism morphisms' isomorphism defined in the afore-mentioned proposition.

$(M_n(\mathbb{H}){)}^x$ is 'groups - homomorphism morphisms' isomorphic to $(f^{″}(M_n(\mathbb{H})){)}^x$ by $g^{\prime }=f{|}_{(M_n(\mathbb{H}){)}^x}:(M_n(\mathbb{H}){)}^x\to (f^{″}(M_n(\mathbb{H})){)}^x$. $GL(n,\mathbb{H})$ is 'groups - homomorphism morphisms' isomorphic to $(f^{″}(M_n(\mathbb{H})){)}^x$ by $g=g^{\prime }\circ h:GL(n,\mathbb{H})\to (f^{″}(M_n(\mathbb{H})){)}^x$. Furthermore, for any $G\in GL(n,\mathbb{H})$ and any $v\in V$, $f^{‴}(h^{\prime }(Gv))=g(G)f^{‴}(h^{\prime }(v))$.

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2: Proof

Let us prove that $g^{\prime }=f{|}_{(M_n(\mathbb{H}){)}^x}:(M_n(\mathbb{H}){)}^x\to (f^{″}(M_n(\mathbb{H})){)}^x$ is a legitimate definition, which means that $f$ maps $(M_n(\mathbb{H}){)}^x$ into $(f^{″}(M_n(\mathbb{H})){)}^x$, which means that for any invertible $M\in (M_n(\mathbb{H}){)}^x$, the determinant of $f(M)$ is nonzero. $M^{-1}M=I$. $f(M^{-1}M)=f(M^{-1})f(M)=f(I)=I$, because $f$ is rings homomorphic. $det(f(M^{-1})f(M))=det(f(M^{-1}))det(f(M))=det(I)=1$. So, $det(f(M))\ne 0$.

$g^{\prime }$ is injective because $f$ is injective.

Before proving that $g^{\prime }$ is surjective, let us prove that $f^{‴}(h^{\prime }(Gv))=g(G)f^{‴}(h^{\prime }(v))$, because that will be used in proving the surjectiveness. That equals $f^{‴}(h(G)h^{\prime }(v))=g^{\prime }(h(G))f^{‴}(h^{\prime }(v))$. The i-th 2 x 2 complex matrix in $f^{‴}(h(G)h^{\prime }(v))$ is $f^{\prime }(\sum _j(h(G{)}_{i,j}{h}^{\prime }(v{)}_j))=\sum _j(f^{\prime }(h(G{)}_{i,j})f^{\prime }(h^{\prime }(v{)}_j)))$, because $f^{\prime }$ is rings homomorphic, which is the i-th 2 x 2 complex matrix in $g^{\prime }(h(G))f^{‴}(h^{\prime }(v))$, by the proposition that the multiplication of any matrix made of any same size blocks by any matrix made of blocks of any multiplicable (with blocks of the former matrix) same size is blocks-wise. So, $f^{‴}(h(G)h^{\prime }(v))=g^{\prime }(h(G))f^{‴}(h^{\prime }(v))$.

Besides, for any $M\in M_n(\mathbb{H})$ and $v\in {\mathbb{H}}^n$, $f^{‴}(Mv)=f(M)f^{‴}(v)$, because the i-th 2 x 2 complex matrix of $f^{‴}(Mv)$ is $f^{\prime }(\sum _j{M}_{i,j}{v}_j)=\sum _j{f}^{\prime }(M_{i,j})f^{\prime }(v_j)$ and the i-th 2 x 2 complex matrix of $f(M)f^{‴}(v)$ is $\sum _j{f}^{\prime }(M_{i,j}{f}^{\prime }(v_j)$, by the proposition that the multiplication of any matrix made of any same size blocks by any matrix made of blocks of any multiplicable (with blocks of the former matrix) same size is blocks-wise and the fact that $f^{\prime }$ is a 'rings - homomorphism morphisms' isomorphism.

Let us prove that $g^{\prime }$ is surjective. For any $f(M)\in (f^{″}(M_n(\mathbb{H})){)}^x$, as the determinant is nonzero, there is the inverse, $(f(M){)}^{-1}$, and $f(M)$ is bijective from ${\mathbb{C}}^{2n}$ to ${\mathbb{C}}^{2n}$. The issue is whether $M$ is bijective from ${\mathbb{H}}^n$ to ${\mathbb{H}}^n$. As for injectiveness, for any $v_1,v_2\in {\mathbb{H}}^n$ such that $v_1\ne v_2$, $M{v}_1\ne M{v}_2$? $f^{‴}(v_1)\ne f^{‴}(v_2)$. $f^{‴}(M{v}_1)=f(M)f^{‴}(v_1)\ne f(M)f^{‴}(v_2)=f^{‴}(M{v}_2)$. So, $M{v}_1\ne M{v}_2$. As for surjectiveness, for any $v^{\prime }\in {\mathbb{H}}^n$, is there $v\in {\mathbb{H}}^n$ such that $v^{\prime }=Mv$? There is the vector, $c\in {\mathbb{C}}^{2n}$, such that $\pi \circ f^{‴}(v^{\prime })=f(M)c$, where $\pi$ is the projection to the 1st column. There is the vector, $v\in {\mathbb{H}}^n$, such that $c=\pi \circ f^{‴}(v)$. Then, $f^{‴}(v^{\prime })=f(M)f^{‴}(v)$, because as $f(M)$ is guaranteed to map $f^{‴}({\mathbb{H}}^n)$ into $f^{‴}({\mathbb{H}}^n)$, $f(M)f^{‴}(v)$ cannot be nothing but $f^{‴}(v^{\prime })$, because as the 1st column of $f^{‴}(v^{\prime })$ is determined from $c$, the 2nd column is determined from the 1st column. Then, $v^{\prime }=Mv$, because if $v^{\prime }\ne Mv$, $f^{‴}(v^{\prime })\ne f^{‴}(Mv)=f(M)f^{‴}(v)$, a contradiction.

So, $g^{\prime }$ is a bijection, and there is the inverse, $g^{\prime -1}$.

$g^{\prime }$ is homomorphic with respect to multiplication, because $f$ is so.

Is $g^{\prime }$ homomorphic with respect to inverse? $g^{\prime }(M^{-1}M)=g^{\prime }(M^{-1})g^{\prime }(M)=g^{\prime }(I)=f(I)=I$. So, $g^{\prime }(M^{-1})=(g^{\prime }(M){)}^{-1}$, so, yes.

So, $g^{\prime }$ is a groups homomorphism.

Is $g^{\prime -1}$ a groups homomorphism?

$g^{\prime -1}$ is homomorphic with respect to multiplication, because $f^{-1}$ is so, while $g^{\prime -1}$ is a restriction of $f^{-1}$.

Is $g^{\prime -1}$ homomorphic with respect to inverse? $g^{\prime -1}((g^{\prime }(M){)}^{-1})g^{\prime -1}(g^{\prime }(M))=g^{\prime -1}((g^{\prime }(M){)}^{-1}{g}^{\prime }(M))=g^{\prime -1}(I)=I$. So, $g^{\prime -1}((g^{\prime }(M){)}^{-1})=(g^{\prime -1}(g^{\prime }(M)){)}^{-1}$, so, yes.

So, $g^{\prime }$ is a 'groups - homomorphism morphisms' isomorphism.

As $h$ and $g^{\prime }$ are 'groups - homomorphism morphisms' isomorphisms, $g=g^{\prime }\circ h$ is a 'groups - homomorphism morphisms' isomorphism.

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References

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