description/proof of that \(n\)-dimensional quaternions general linear group is 'groups - homomorphisms' isomorphic to group of nonzero determinant corresponding \(2 n \times 2 n\) complex matrices and can be represented by latter
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of quaternions division associative algebra.
- The reader knows a definition of general linear group of module.
- The reader admits the proposition that the quaternions division associative algebra is 'rings - homomorphisms' isomorphic to a subring of the \(2 \times 2\) complex matrices ring.
- The reader admits the proposition that the ring of the \(n \times n\) quaternion matrices is 'rings - homomorphisms' isomorphic to a subring of the \(2 n \times 2 n\) complex matrices, via the between-the-quaternions-division-associative-algebra-and-a-subring-of-\(2 \times 2\)-complex-matrices 'rings - homomorphisms' isomorphism.
- The reader admits the proposition that for any module with any \(d\)-elements basis and the general linear group of the module, there is the canonical 'groups - homomorphisms' isomorphism with respect to the basis from the general linear group onto the group of the invertible \(d \times d\) ring matrices.
- The reader admits the proposition that any ring is canonically a module with a \(1\)-element basis.
- The reader admits the proposition that any finite product module of modules with bases has this basis.
- The reader admits the proposition that for any ring, the multiplications of any matrices over the ring are associative.
- The reader admits the proposition that for any map from any module with any \(d_1\)-elements basis into any module with any \(d_2\)-elements basis over any same ring, the map is linear if and only if the map is represented by the \(d_1 \times d_2\) ring matrix.
- 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.
- The reader admits the proposition that for any linear map between modules with any same cardinality bases, the map is a bijection if and only if the corresponding matrix with respect to the bases is invertible.
- The reader admits the proposition that any map between any groups that maps the product of any 2 elements to the product of the images of the elements is a group homomorphism.
- The reader admits the proposition that any bijective group homomorphism is a 'groups - homomorphisms' isomorphism.
Target Context
- The reader will have a description and a proof of the proposition that the \(n\)-dimensional quaternions general linear group is 'groups - homomorphisms' isomorphic to the group of the nonzero determinant corresponding \(2 n \times 2 n\) 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.
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.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(\mathbb{H}\): \(= \text{ the quaternions division associative algebra }\), \(\in \{\text{ the rings }\}\), \(= \text{ the canonical } \mathbb{H} \text{ module with the basis }, \{1\}\)
\(\mathbb{H}^n\): \(= \text{ the product module with the basis }, B := \{b^1 = (1, 0, ..., 0), ..., b^n = (0, ..., 0, 1)\}\)
\(GL (\mathbb{H}^n)\): \(= \text{ the general linear group of } \mathbb{H}^n\)
\(M_n (\mathbb{H})\): \(= \text{ the ring of the } n \times n \mathbb{H} \text{ matrices }\)
\(M_n (\mathbb{H})^{\times}\): \(= \text{ the group of the invertible elements of } M_n (\mathbb{H})\)
\(f\): \(: \mathbb{H} \to R\), \(= \text{ the 'rings - homomorphisms' isomorphism }\) defined in the proposition that the quaternions division associative algebra is 'rings - homomorphisms' isomorphic to a subring of the \(2 \times 2\) complex matrices ring
\(f''\): \(: M_n (\mathbb{H}) \to f' (M_n (\mathbb{H}))\), \(= \text{ the 'rings - homomorphisms' isomorphism }\) defined in the proposition that the ring of the \(n \times n\) quaternion matrices is 'rings - homomorphisms' isomorphic to a subring of the \(2 n \times 2 n\) complex matrices, via the between-the-quaternions-division-associative-algebra-and-a-subring-of-\(2 \times 2\)-complex-matrices 'rings - homomorphisms' isomorphism
\(f'' (M_n (\mathbb{H}))^{\times}\): \(= \text{ the group of the nonzero-determinant elements of } f'' (M_n (\mathbb{H}))\)
\(g\): \(: GL (\mathbb{H}^n) \to M_n (\mathbb{H})^{\times}\), \(= \text{ the 'groups - homomorphisms' isomorphism with respect to } B\) defined in the proposition that for any module with any \(d\)-elements basis and the general linear group of the module, there is the canonical 'groups - homomorphisms' isomorphism with respect to the basis from the general linear group onto the group of the invertible \(d \times d\) ring matrices
\(f'''\): \(: \mathbb{H}^n \to \mathbb{C}^{2 \times 2 n}\) which replaces each component of each \(p \in \mathbb{H}^n\) regarded as the row with \(f (p_j)\)
\(g'\): \(: M_n (\mathbb{H})^{\times} \to f'' (M_n (\mathbb{H}))^{\times}\), \(= f'' \vert_{M_n (\mathbb{H})^{\times}}\)
\(g''\): \(: GL (\mathbb{H}^n) \to f'' (M_n (\mathbb{H}))^{\times}\), \(= g' \circ g\)
//
Statements:
\(g' \in \{ \text{ the 'groups - homomorphisms' isomorphisms } \}\)
\(\land\)
\(g'' \in \{ \text{ the 'groups - homomorphisms' isomorphisms } \}\)
\(\land\)
\(\forall h \in GL (\mathbb{H}^n), \forall m \in \mathbb{H}^n (f''' (h (m)) = f''' (m) g'' (h))\)
//
2: Proof
Whole Strategy: Step 1: see that \(\mathbb{H}^n\) is a module with \(B\), that \(M_n (\mathbb{H})\) is a ring, and that \(M_n (\mathbb{H})^{\times}\) is a group; Step 2: see that \(g'\) is valid; Step 3: see that \(g'\) is a bijection; Step 4: see that \(g'\) is a group homomorphism; Step 5: conclude that \(g'\) and \(g''\) are some 'groups - homomorphisms' isomorphisms; Step 6: see that \(f''' (h (m)) = f''' (m) g'' (h)\).
Step 1:
Let us see that what are listed in Entities are really as are claimed.
\(\mathbb{H}\) is a ring, as is mentioned in Note for the definition of quaternions division associative algebra.
\(\mathbb{H}\) is an \(\mathbb{H}\) module with the basis, \(\{1\}\), by the proposition that any ring is canonically a module with a \(1\)-element basis.
\(\mathbb{H}^n\) has the basis, \(B\), by the proposition that any finite product module of modules with bases has this basis.
\(M_n (\mathbb{H})\) is a ring, by Note for the definition of ring of \(n \times n\) ring matrices.
\(M_n (\mathbb{H})^{\times}\) is a group, by the proposition that for any module with any \(d\)-elements basis and the general linear group of the module, there is the canonical 'groups - homomorphisms' isomorphism with respect to the basis from the general linear group onto the group of the invertible \(d \times d\) ring matrices.
We will see later that \(f'' (M_n (\mathbb{H}))^{\times}\) is a group.
Step 2:
Let us see that \(g'\) is valid.
Let \(m \in M_n (\mathbb{H})^{\times}\) be any.
There is the \(m^{- 1} \in M_n (\mathbb{H})^{\times}\), because \(M_n (\mathbb{H})^{\times}\) is a group, and \(f'' (I) = f'' (m^{- 1} m) = f'' (m^{- 1}) f'' (m)\), because \(f''\) is a rings homomorphism, but \(f'' (I) = I\), because \(f''\) is a rings homomorphism, so, \(f'' (m^{- 1}) f'' (m) = I\), and \(det (f'' (m^{- 1}) f'' (m)) = det I = 1\), but the left hand side is \(det f'' (m^{- 1}) det f'' (m)\), as is well known, so, \(det f'' (m) \neq 0\).
So, \(f'' (m) \in f'' (M_n (\mathbb{H}))^{\times}\).
So, \(g'\) is into \(f'' (M_n (\mathbb{H}))^{\times}\).
So, \(g'\) is valid.
Step 3:
\(g'\) is an injection, because \(f''\) is so, because \(f''\) is a 'rings - homomorphisms' isomorphism.
There is the bijective \(\widetilde{g}: End (\mathbb{H}^n) \to M_n (\mathbb{H})\) where \(End (\mathbb{H}^n)\) is the set of the module endomorphisms on \(\mathbb{H}^n\), by the proposition that for any map from any module with any \(d_1\)-elements basis into any module with any \(d_2\)-elements basis over any same ring, the map is linear if and only if the map is represented by the \(d_1 \times d_2\) ring matrix (\(g\) is the restriction of \(\widetilde{g}\)).
Let us see that for each \(h \in End (\mathbb{H}^n)\) and \(m \in \mathbb{H}^n\), \(f''' (h (m)) = f''' (m) f'' \circ \widetilde{g} (h)\).
The \(j\)-th \(2 \times 2\) block in \(f''' (h (m))\) is \(f (h (m)_j) = f (m_l \widetilde{g} (h)^l_j) = f (m_l) f (\widetilde{g} (h)^l_j)\), because \(f\) is a ring homomorphism.
The \(j\)-th \(2 \times 2\) block in \(f''' (m) f'' \circ \widetilde{g} (h)\) is \(f (m_l) f (\widetilde{g} (h)^l_j)\), 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.
So, \(f''' (h (m)) = f''' (m) f'' \circ \widetilde{g} (h)\).
Let us see that \(g'\) is a surjection.
For each \(f'' (M) \in f'' (M_n (\mathbb{H}))^{\times}\), as the determinant is nonzero, there is the inverse, \(f'' (M)^{-1} \in M_{2 n} (\mathbb{C})\), and \(f'' (M)\) represents the bijection from \(\mathbb{C}^{2 n}\) to \(\mathbb{C}^{2 n}\), by the proposition that for any linear map between modules with any same cardinality bases, the map is a bijection if and only if the corresponding matrix with respect to the bases is invertible: \(\mathbb{C}^{2 n}\) is a module with the canonical \(2 n\)-elements basis, by the proposition that any ring is canonically a module with a \(1\)-element basis and the proposition that any finite product module of modules with bases has this basis.
\(\widetilde{g}^{- 1} (M) \in End (\mathbb{H}^n)\).
The issue is whether \(\widetilde{g}^{- 1} (M)\) is a bijection.
\(\widetilde{g}^{- 1} (M)\) is an injection, because for each \(m, m' \in \mathbb{H}^n\) such that \(m \neq m'\), \(f''' (m) \neq f''' (m')\), and \(f''' (\widetilde{g}^{- 1} (M) (m)) = f''' (m) f'' \circ \widetilde{g} (\widetilde{g}^{- 1} (M)) = f''' (m) f'' (M) \neq f''' (m') f'' (M) = f''' (m') f'' \circ \widetilde{g} (\widetilde{g}^{- 1} (M)) = f''' (\widetilde{g}^{- 1} (M) (m'))\), because \(f'' (M)\) represents a bijection, so, \(\widetilde{g}^{- 1} (M) (m) \neq \widetilde{g}^{- 1} (M) (m')\).
Let us see that \(\widetilde{g}^{- 1} (M)\) is a surjection.
Let \(m' \in \mathbb{H}^n\) be any.
There is \(v \in \mathbb{C}^{2 n}\) such that \(\pi_1 \circ f''' (m') = v f'' (M)\), where \(\pi_1\) is the projection to the 1st row, because \(f'' (M)\) represents a bijection.
There is \(m \in \mathbb{H}^n\) such that \(v = \pi_1 \circ f''' (m)\), because when the \(j\)-th \(1 \times 2\) block of \(v\) is \((a + b i, c + d i)\), \(m_j\) can be taken to be \(a + b i + c j + d k\).
Then, \(f''' (m') = f''' (m) f'' (M)\), because while \(\pi_1 \circ f''' (m') = v f'' (M) = (\pi_1 \circ f''' (m)) f'' (M) = \pi_1 (f''' (m) f'' (M))\), which is directly from the definition of matrix multiplication, \(f''' (m) f'' (M) = f''' (m) f'' \circ \widetilde{g} (\widetilde{g}^{- 1} (M)) = f''' (\widetilde{g}^{- 1} (M) (m))\), which has shown that \(f''' (m) f'' (M)\) is a value of \(f'''\), so, \(f''' (m) f'' (M)\) cannot be nothing but \(f''' (m')\), because for any value of \(f'''\), the 2nd row is uniquely determined from the 1st row, so, when the 1st rows of any \(2\) values of \(f'''\) are equal, the 2nd rows of the values are equal.
Then, \(m' = \widetilde{g}^{- 1} (M) (m)\), because if \(m' \neq \widetilde{g}^{- 1} (M) (m)\), \(f''' (m') \neq f''' (\widetilde{g}^{- 1} (M) (m)) = f''' (m) f'' \circ \widetilde{g} (\widetilde{g}^{- 1} (M)) = f''' (m) f'' (M)\), a contradiction against \(f''' (m') = f''' (m) f'' (M)\).
So, \(\widetilde{g}^{- 1} (M)\) is a surjection.
So, \(\widetilde{g}^{- 1} (M)\) is a bijection.
So, \(M\) is invertible, by the proposition that for any linear map between modules with any same cardinality bases, the map is a bijection if and only if the corresponding matrix with respect to the bases is invertible.
So, \(M \in M_n (\mathbb{H})^{\times}\).
So, \(g'\) is a surjection.
So, \(g'\) is a bijection.
Step 4:
Let us see that \(f'' (M_n (\mathbb{H}))^{\times}\) is indeed a group.
For each \(f'' (m), f'' (m') \in f'' (M_n (\mathbb{H}))^{\times}\), \(f'' (m) f'' (m') = f'' (m m') \in f'' (M_n (\mathbb{H}))\), because \(f''\) is a ring homomorphism, and \(det (f'' (m) f'' (m')) = det f'' (m) det f'' (m')\), as is well known, \(\neq 0\), so, \(f'' (m) f'' (m') \in f'' (M_n (\mathbb{H}))^{\times}\), so, \(f'' (M_n (\mathbb{H}))^{\times}\) is closed under multiplication.
The multiplications are associative, by the proposition that for any ring, the multiplications of any matrices over the ring are associative.
\(f'' (I) = I\), because \(f''\) is a rings homomorphism, which is in \(f'' (M_n (\mathbb{H}))^{\times}\), because it has the nonzero determinant, which is the identity element of \(f'' (M_n (\mathbb{H}))^{\times}\)
For each \(f'' (m) \in f'' (M_n (\mathbb{H}))^{\times}\), as \(g'\) is a bijection, by Step 3, \(m \in M_n (\mathbb{H})^{\times}\), so, there is an inverse, \(m' \in M_n (\mathbb{H})^{\times}\), such that \(m m' = I\) and \(m' m = I\), so, \(I = f'' (I) = f'' (m m') = f'' (m) f'' (m')\) and \(I = f'' (I) = f'' (m' m) = f'' (m') f'' (m)\), because \(f''\) is a ring homomorphism, so, \(f'' (m)^{- 1} = f'' (m') \in f'' (M_n (\mathbb{H}))^{\times}\).
So, \(f'' (M_n (\mathbb{H}))^{\times}\) is a group.
Let us see that \(g'\) is a group homomorphism.
For each \(M, M' \in M_n (\mathbb{H})^{\times}\), \(g' (M M') = f'' (M M') = f'' (M) f'' (M') = g' (M) g' (M')\), because \(f''\) is a ring homomorphism.
\(g'\) is a group homomorphism, by the proposition that any map between any groups that maps the product of any 2 elements to the product of the images of the elements is a group homomorphism.
Step 5:
\(g'\) is a 'groups - homomorphisms' isomorphism, by the proposition that any bijective group homomorphism is a 'groups - homomorphisms' isomorphism.
\(g'' = g' \circ g\) is a 'groups - homomorphisms' isomorphism, because \(g\) and \(g'\) are some 'groups - homomorphisms' isomorphisms.
Step 6:
\(\forall h \in GL (\mathbb{H}^n), \forall m \in \mathbb{H}^n (f''' (h (m)) = f''' (m) g'' (h))\), is just a restriction of \(f''' (h (m)) = f''' (m) f'' \circ \widetilde{g} (h)\) seen in Step 3, because \(f'' \circ \widetilde{g} (h) = f'' \circ g (h) = g' \circ g (h) = g'' (h)\).
