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proof of basic fact that torus actions are diagonalizable
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Suppose a torus $T=(mathbbC^ast)^n$ acts on a finite dimensional vector space $W$, and define for $m in M$ ($M$ is the character lattice of $T$) the eigenspace $W_m$ by
$$W_m = w in W mid tcdot w = chi^m(t)w text for all t in T $$
i.e. for $w in W_m$ is a simultaneous eigenvector for all $t in T$, with eigenvalue $chi^m(t)$ depending on $t in T$. Then it is a famous fact $$W=undersetm in M bigoplus W_m$$
Can someone provide a somewhat self-contained proof of this result? I don't know much about the theory of algebraic groups.
linear-algebra representation-theory algebraic-groups
asked Jun 9 '14 at 20:04
ykm
916412
add a comment |Â
up vote
3
down vote
favorite
Suppose a torus $T=(mathbbC^ast)^n$ acts on a finite dimensional vector space $W$, and define for $m in M$ ($M$ is the character lattice of $T$) the eigenspace $W_m$ by
$$W_m = w in W mid tcdot w = chi^m(t)w text for all t in T $$
i.e. for $w in W_m$ is a simultaneous eigenvector for all $t in T$, with eigenvalue $chi^m(t)$ depending on $t in T$. Then it is a famous fact $$W=undersetm in M bigoplus W_m$$
Can someone provide a somewhat self-contained proof of this result? I don't know much about the theory of algebraic groups.
linear-algebra representation-theory algebraic-groups
asked Jun 9 '14 at 20:04
ykm
916412
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Suppose a torus $T=(mathbbC^ast)^n$ acts on a finite dimensional vector space $W$, and define for $m in M$ ($M$ is the character lattice of $T$) the eigenspace $W_m$ by
$$W_m = w in W mid tcdot w = chi^m(t)w text for all t in T $$
i.e. for $w in W_m$ is a simultaneous eigenvector for all $t in T$, with eigenvalue $chi^m(t)$ depending on $t in T$. Then it is a famous fact $$W=undersetm in M bigoplus W_m$$
Can someone provide a somewhat self-contained proof of this result? I don't know much about the theory of algebraic groups.
linear-algebra representation-theory algebraic-groups
asked Jun 9 '14 at 20:04
ykm
916412
Suppose a torus $T=(mathbbC^ast)^n$ acts on a finite dimensional vector space $W$, and define for $m in M$ ($M$ is the character lattice of $T$) the eigenspace $W_m$ by
$$W_m = w in W mid tcdot w = chi^m(t)w text for all t in T $$
i.e. for $w in W_m$ is a simultaneous eigenvector for all $t in T$, with eigenvalue $chi^m(t)$ depending on $t in T$. Then it is a famous fact $$W=undersetm in M bigoplus W_m$$
Can someone provide a somewhat self-contained proof of this result? I don't know much about the theory of algebraic groups.
linear-algebra representation-theory algebraic-groups
asked Jun 9 '14 at 20:04
ykm
916412
asked Jun 9 '14 at 20:04
ykm
916412
asked Jun 9 '14 at 20:04
ykm
916412
asked Jun 9 '14 at 20:04
ykm
916412
916412
add a comment |Â
add a comment |Â
1 Answer
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$defCmathbbC$
My answer uses results from James Humphreys' Linear Algebraic Groups. I'll cite the pages which have the results I'm going to use because the proofs he gives are entirely self-contained and I would mostly be transcribing them otherwise.
The key idea is that the torus $(mathbbC^*)^n$ consists of commuting semisimple elements and that homomorphisms of algebraic groups preserve semisimplicity. This is all you need to know if you don't want to read the very long answer that follows.
For an algebraic group $G$ and an element $x in G$, we say that $x$ is semisimple if the action of $x$ on the coordinate ring $mathbbC[G]$ defined by $(x cdot f(g)) = f(gcdot x)$ can be diagonalized. Now, the coordinate ring of $(mathbbC^*)^n$ is $mathbbC[x_1^pm, ldots, x_n^pm].$ Now, if $r = (r_1, ldots, r_n)$ is an element of $(mathbbC^*)^n$ and $g$ is an arbitrary element of the same, then for any $a_1, ldots, a_n$,
$$(rcdot x_1^a_1cdots x_n^a_n)(g) = r_1^a_1cdots r_n^a_n(x_1^a_1cdots x_n^a_n)(g).$$
Hence
$$r cdot x_1^a_1 cdots x_n^a_n = r_1^a_1cdots r_n^a_n (x_1^a_1cdots x_n^a_n).$$
Hence, each of the monomials in the coordinate ring of $(mathbbC^*)^n$ is an eigenvector for $r$. Thus, every element in $(mathbbC^*)^n$ acts diagonally on the coordinate ring and is hence semisimple.
The next result is found in page 99 of Humphreys, where he proves (among other things) that homomorphisms of algebraic groups preserve semisimplicty. That is, if $rho: G rightarrow G'$ is a morphism of algebraic groups and if $g in G$ is semisimple, then $rho(g)$ is semisimple in $G'.$
So now, if we have a representation $V$ of $(C^*)^n$, then we have a homomorphism $rho: (C^*)^n rightarrow GL(V)$ whose image consists entirely of semisimple elements. So now, we show that if an element $g in GL(V)$ is semisimple then it acts diagonally on $V$. This will then imply that each element of $rho((C^*)^n)$ acts diagonally on $V$ as and hence the entire action can be simultaneously diagonalized since the operators also commute with each other.
To finish the proof, we use the multiplicative Jordan decomposition. This can be found on page 96 of Humphreys (and many other sources I would imagine.) This says that for any element $g in GL(V)$, there exist $g_s, g_u in GL(V)$ such that
- $g, g_s, g_u$ commute with any linear map $Vrightarrow V$ that commutes with $g$.
- $g_s$ is diagonalizable.
- $g_u$ is unipotent i.e. there exists some $m$ such that $(g_u - 1)^m = 0.$
So now, since $g_s$ is diagonalizable, it must act semisimply on $C[GL(V)]$. This is because we let $x_i, j$ be coordinates for $GL(V) cong M(n times n)$ in the basis such that $g_s$ is diagonal, then $C[GL(V)] = C[X_1,1, ldots, x_n, n]_det$ and the elements of the form
$$fracx_1, 1^a_1, 1cdots x_n, n^a_n, ndet^l$$
form a spanning set in $C[GL(V)]$ of eigenvectors for $g_s$ and hence for $g_s^-1.$ As $g$ also acts diagonally on $C[GL(V)]$, and commutes with $g_s^-1$, the action of $g$ and $g_s^-1$ on $C[GL(V)]$ must be simulataneously diagonalizable. Hence, $gg_s^-1 = g_u$ must also act diagonally on $C[GL(V)]$.
But $g_u - 1$ acts nilpotently on $C[GL(V)]$ because its $m$th power is $0$. Hence, the action of $g_u$ in $C[GL(V)]$ must have eigenvalue $1$ and is hence the identity. Since the representation of an algebraic group in its coordinate ring is faithful, this implies that $g_u = 1$ and hence $g = g_s$ is diagonalizable in $GL(V)$. This is what we needed.
answered Jun 10 '14 at 7:43
Siddharth Venkatesh
1,97565
This is useful, thank you. I had one question. I know that a finite set of commuting diagonalizable linear operators can be simultaneously diagonalized. But $(mathbbC^ast)^n$ has infinitely many elements, so what is the argument that they can all be simultaneously diagonilized?
– ykm
Jun 11 '14 at 4:57
$(mathbbC^*)^n$ is the Zariski closure of the subgroup generated by finitely many elements. If these finitely many elements are diagonalized, so is the subgroup they generate and hence so is its Zariski closure as the set of diagonal matrices is Zariski closed.
– Siddharth Venkatesh
Jun 11 '14 at 8:08
add a comment |Â
1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
$defCmathbbC$
My answer uses results from James Humphreys' Linear Algebraic Groups. I'll cite the pages which have the results I'm going to use because the proofs he gives are entirely self-contained and I would mostly be transcribing them otherwise.
The key idea is that the torus $(mathbbC^*)^n$ consists of commuting semisimple elements and that homomorphisms of algebraic groups preserve semisimplicity. This is all you need to know if you don't want to read the very long answer that follows.
For an algebraic group $G$ and an element $x in G$, we say that $x$ is semisimple if the action of $x$ on the coordinate ring $mathbbC[G]$ defined by $(x cdot f(g)) = f(gcdot x)$ can be diagonalized. Now, the coordinate ring of $(mathbbC^*)^n$ is $mathbbC[x_1^pm, ldots, x_n^pm].$ Now, if $r = (r_1, ldots, r_n)$ is an element of $(mathbbC^*)^n$ and $g$ is an arbitrary element of the same, then for any $a_1, ldots, a_n$,
$$(rcdot x_1^a_1cdots x_n^a_n)(g) = r_1^a_1cdots r_n^a_n(x_1^a_1cdots x_n^a_n)(g).$$
Hence
$$r cdot x_1^a_1 cdots x_n^a_n = r_1^a_1cdots r_n^a_n (x_1^a_1cdots x_n^a_n).$$
Hence, each of the monomials in the coordinate ring of $(mathbbC^*)^n$ is an eigenvector for $r$. Thus, every element in $(mathbbC^*)^n$ acts diagonally on the coordinate ring and is hence semisimple.
The next result is found in page 99 of Humphreys, where he proves (among other things) that homomorphisms of algebraic groups preserve semisimplicty. That is, if $rho: G rightarrow G'$ is a morphism of algebraic groups and if $g in G$ is semisimple, then $rho(g)$ is semisimple in $G'.$
So now, if we have a representation $V$ of $(C^*)^n$, then we have a homomorphism $rho: (C^*)^n rightarrow GL(V)$ whose image consists entirely of semisimple elements. So now, we show that if an element $g in GL(V)$ is semisimple then it acts diagonally on $V$. This will then imply that each element of $rho((C^*)^n)$ acts diagonally on $V$ as and hence the entire action can be simultaneously diagonalized since the operators also commute with each other.
To finish the proof, we use the multiplicative Jordan decomposition. This can be found on page 96 of Humphreys (and many other sources I would imagine.) This says that for any element $g in GL(V)$, there exist $g_s, g_u in GL(V)$ such that
- $g, g_s, g_u$ commute with any linear map $Vrightarrow V$ that commutes with $g$.
- $g_s$ is diagonalizable.
- $g_u$ is unipotent i.e. there exists some $m$ such that $(g_u - 1)^m = 0.$
So now, since $g_s$ is diagonalizable, it must act semisimply on $C[GL(V)]$. This is because we let $x_i, j$ be coordinates for $GL(V) cong M(n times n)$ in the basis such that $g_s$ is diagonal, then $C[GL(V)] = C[X_1,1, ldots, x_n, n]_det$ and the elements of the form
$$fracx_1, 1^a_1, 1cdots x_n, n^a_n, ndet^l$$
form a spanning set in $C[GL(V)]$ of eigenvectors for $g_s$ and hence for $g_s^-1.$ As $g$ also acts diagonally on $C[GL(V)]$, and commutes with $g_s^-1$, the action of $g$ and $g_s^-1$ on $C[GL(V)]$ must be simulataneously diagonalizable. Hence, $gg_s^-1 = g_u$ must also act diagonally on $C[GL(V)]$.
But $g_u - 1$ acts nilpotently on $C[GL(V)]$ because its $m$th power is $0$. Hence, the action of $g_u$ in $C[GL(V)]$ must have eigenvalue $1$ and is hence the identity. Since the representation of an algebraic group in its coordinate ring is faithful, this implies that $g_u = 1$ and hence $g = g_s$ is diagonalizable in $GL(V)$. This is what we needed.
answered Jun 10 '14 at 7:43
Siddharth Venkatesh
1,97565
This is useful, thank you. I had one question. I know that a finite set of commuting diagonalizable linear operators can be simultaneously diagonalized. But $(mathbbC^ast)^n$ has infinitely many elements, so what is the argument that they can all be simultaneously diagonilized?
– ykm
Jun 11 '14 at 4:57
$(mathbbC^*)^n$ is the Zariski closure of the subgroup generated by finitely many elements. If these finitely many elements are diagonalized, so is the subgroup they generate and hence so is its Zariski closure as the set of diagonal matrices is Zariski closed.
– Siddharth Venkatesh
Jun 11 '14 at 8:08
add a comment |Â
up vote
8
down vote
accepted
$defCmathbbC$
My answer uses results from James Humphreys' Linear Algebraic Groups. I'll cite the pages which have the results I'm going to use because the proofs he gives are entirely self-contained and I would mostly be transcribing them otherwise.
The key idea is that the torus $(mathbbC^*)^n$ consists of commuting semisimple elements and that homomorphisms of algebraic groups preserve semisimplicity. This is all you need to know if you don't want to read the very long answer that follows.
For an algebraic group $G$ and an element $x in G$, we say that $x$ is semisimple if the action of $x$ on the coordinate ring $mathbbC[G]$ defined by $(x cdot f(g)) = f(gcdot x)$ can be diagonalized. Now, the coordinate ring of $(mathbbC^*)^n$ is $mathbbC[x_1^pm, ldots, x_n^pm].$ Now, if $r = (r_1, ldots, r_n)$ is an element of $(mathbbC^*)^n$ and $g$ is an arbitrary element of the same, then for any $a_1, ldots, a_n$,
$$(rcdot x_1^a_1cdots x_n^a_n)(g) = r_1^a_1cdots r_n^a_n(x_1^a_1cdots x_n^a_n)(g).$$
Hence
$$r cdot x_1^a_1 cdots x_n^a_n = r_1^a_1cdots r_n^a_n (x_1^a_1cdots x_n^a_n).$$
Hence, each of the monomials in the coordinate ring of $(mathbbC^*)^n$ is an eigenvector for $r$. Thus, every element in $(mathbbC^*)^n$ acts diagonally on the coordinate ring and is hence semisimple.
The next result is found in page 99 of Humphreys, where he proves (among other things) that homomorphisms of algebraic groups preserve semisimplicty. That is, if $rho: G rightarrow G'$ is a morphism of algebraic groups and if $g in G$ is semisimple, then $rho(g)$ is semisimple in $G'.$
So now, if we have a representation $V$ of $(C^*)^n$, then we have a homomorphism $rho: (C^*)^n rightarrow GL(V)$ whose image consists entirely of semisimple elements. So now, we show that if an element $g in GL(V)$ is semisimple then it acts diagonally on $V$. This will then imply that each element of $rho((C^*)^n)$ acts diagonally on $V$ as and hence the entire action can be simultaneously diagonalized since the operators also commute with each other.
To finish the proof, we use the multiplicative Jordan decomposition. This can be found on page 96 of Humphreys (and many other sources I would imagine.) This says that for any element $g in GL(V)$, there exist $g_s, g_u in GL(V)$ such that
- $g, g_s, g_u$ commute with any linear map $Vrightarrow V$ that commutes with $g$.
- $g_s$ is diagonalizable.
- $g_u$ is unipotent i.e. there exists some $m$ such that $(g_u - 1)^m = 0.$
So now, since $g_s$ is diagonalizable, it must act semisimply on $C[GL(V)]$. This is because we let $x_i, j$ be coordinates for $GL(V) cong M(n times n)$ in the basis such that $g_s$ is diagonal, then $C[GL(V)] = C[X_1,1, ldots, x_n, n]_det$ and the elements of the form
$$fracx_1, 1^a_1, 1cdots x_n, n^a_n, ndet^l$$
form a spanning set in $C[GL(V)]$ of eigenvectors for $g_s$ and hence for $g_s^-1.$ As $g$ also acts diagonally on $C[GL(V)]$, and commutes with $g_s^-1$, the action of $g$ and $g_s^-1$ on $C[GL(V)]$ must be simulataneously diagonalizable. Hence, $gg_s^-1 = g_u$ must also act diagonally on $C[GL(V)]$.
But $g_u - 1$ acts nilpotently on $C[GL(V)]$ because its $m$th power is $0$. Hence, the action of $g_u$ in $C[GL(V)]$ must have eigenvalue $1$ and is hence the identity. Since the representation of an algebraic group in its coordinate ring is faithful, this implies that $g_u = 1$ and hence $g = g_s$ is diagonalizable in $GL(V)$. This is what we needed.
answered Jun 10 '14 at 7:43
Siddharth Venkatesh
1,97565
This is useful, thank you. I had one question. I know that a finite set of commuting diagonalizable linear operators can be simultaneously diagonalized. But $(mathbbC^ast)^n$ has infinitely many elements, so what is the argument that they can all be simultaneously diagonilized?
– ykm
Jun 11 '14 at 4:57
$(mathbbC^*)^n$ is the Zariski closure of the subgroup generated by finitely many elements. If these finitely many elements are diagonalized, so is the subgroup they generate and hence so is its Zariski closure as the set of diagonal matrices is Zariski closed.
– Siddharth Venkatesh
Jun 11 '14 at 8:08
add a comment |Â
up vote
8
down vote
accepted
up vote
8
down vote
accepted
$defCmathbbC$
My answer uses results from James Humphreys' Linear Algebraic Groups. I'll cite the pages which have the results I'm going to use because the proofs he gives are entirely self-contained and I would mostly be transcribing them otherwise.
The key idea is that the torus $(mathbbC^*)^n$ consists of commuting semisimple elements and that homomorphisms of algebraic groups preserve semisimplicity. This is all you need to know if you don't want to read the very long answer that follows.
For an algebraic group $G$ and an element $x in G$, we say that $x$ is semisimple if the action of $x$ on the coordinate ring $mathbbC[G]$ defined by $(x cdot f(g)) = f(gcdot x)$ can be diagonalized. Now, the coordinate ring of $(mathbbC^*)^n$ is $mathbbC[x_1^pm, ldots, x_n^pm].$ Now, if $r = (r_1, ldots, r_n)$ is an element of $(mathbbC^*)^n$ and $g$ is an arbitrary element of the same, then for any $a_1, ldots, a_n$,
$$(rcdot x_1^a_1cdots x_n^a_n)(g) = r_1^a_1cdots r_n^a_n(x_1^a_1cdots x_n^a_n)(g).$$
Hence
$$r cdot x_1^a_1 cdots x_n^a_n = r_1^a_1cdots r_n^a_n (x_1^a_1cdots x_n^a_n).$$
Hence, each of the monomials in the coordinate ring of $(mathbbC^*)^n$ is an eigenvector for $r$. Thus, every element in $(mathbbC^*)^n$ acts diagonally on the coordinate ring and is hence semisimple.
The next result is found in page 99 of Humphreys, where he proves (among other things) that homomorphisms of algebraic groups preserve semisimplicty. That is, if $rho: G rightarrow G'$ is a morphism of algebraic groups and if $g in G$ is semisimple, then $rho(g)$ is semisimple in $G'.$
So now, if we have a representation $V$ of $(C^*)^n$, then we have a homomorphism $rho: (C^*)^n rightarrow GL(V)$ whose image consists entirely of semisimple elements. So now, we show that if an element $g in GL(V)$ is semisimple then it acts diagonally on $V$. This will then imply that each element of $rho((C^*)^n)$ acts diagonally on $V$ as and hence the entire action can be simultaneously diagonalized since the operators also commute with each other.
To finish the proof, we use the multiplicative Jordan decomposition. This can be found on page 96 of Humphreys (and many other sources I would imagine.) This says that for any element $g in GL(V)$, there exist $g_s, g_u in GL(V)$ such that
- $g, g_s, g_u$ commute with any linear map $Vrightarrow V$ that commutes with $g$.
- $g_s$ is diagonalizable.
- $g_u$ is unipotent i.e. there exists some $m$ such that $(g_u - 1)^m = 0.$
So now, since $g_s$ is diagonalizable, it must act semisimply on $C[GL(V)]$. This is because we let $x_i, j$ be coordinates for $GL(V) cong M(n times n)$ in the basis such that $g_s$ is diagonal, then $C[GL(V)] = C[X_1,1, ldots, x_n, n]_det$ and the elements of the form
$$fracx_1, 1^a_1, 1cdots x_n, n^a_n, ndet^l$$
form a spanning set in $C[GL(V)]$ of eigenvectors for $g_s$ and hence for $g_s^-1.$ As $g$ also acts diagonally on $C[GL(V)]$, and commutes with $g_s^-1$, the action of $g$ and $g_s^-1$ on $C[GL(V)]$ must be simulataneously diagonalizable. Hence, $gg_s^-1 = g_u$ must also act diagonally on $C[GL(V)]$.
But $g_u - 1$ acts nilpotently on $C[GL(V)]$ because its $m$th power is $0$. Hence, the action of $g_u$ in $C[GL(V)]$ must have eigenvalue $1$ and is hence the identity. Since the representation of an algebraic group in its coordinate ring is faithful, this implies that $g_u = 1$ and hence $g = g_s$ is diagonalizable in $GL(V)$. This is what we needed.
answered Jun 10 '14 at 7:43
Siddharth Venkatesh
1,97565
$defCmathbbC$
My answer uses results from James Humphreys' Linear Algebraic Groups. I'll cite the pages which have the results I'm going to use because the proofs he gives are entirely self-contained and I would mostly be transcribing them otherwise.
The key idea is that the torus $(mathbbC^*)^n$ consists of commuting semisimple elements and that homomorphisms of algebraic groups preserve semisimplicity. This is all you need to know if you don't want to read the very long answer that follows.
For an algebraic group $G$ and an element $x in G$, we say that $x$ is semisimple if the action of $x$ on the coordinate ring $mathbbC[G]$ defined by $(x cdot f(g)) = f(gcdot x)$ can be diagonalized. Now, the coordinate ring of $(mathbbC^*)^n$ is $mathbbC[x_1^pm, ldots, x_n^pm].$ Now, if $r = (r_1, ldots, r_n)$ is an element of $(mathbbC^*)^n$ and $g$ is an arbitrary element of the same, then for any $a_1, ldots, a_n$,
$$(rcdot x_1^a_1cdots x_n^a_n)(g) = r_1^a_1cdots r_n^a_n(x_1^a_1cdots x_n^a_n)(g).$$
Hence
$$r cdot x_1^a_1 cdots x_n^a_n = r_1^a_1cdots r_n^a_n (x_1^a_1cdots x_n^a_n).$$
Hence, each of the monomials in the coordinate ring of $(mathbbC^*)^n$ is an eigenvector for $r$. Thus, every element in $(mathbbC^*)^n$ acts diagonally on the coordinate ring and is hence semisimple.
The next result is found in page 99 of Humphreys, where he proves (among other things) that homomorphisms of algebraic groups preserve semisimplicty. That is, if $rho: G rightarrow G'$ is a morphism of algebraic groups and if $g in G$ is semisimple, then $rho(g)$ is semisimple in $G'.$
So now, if we have a representation $V$ of $(C^*)^n$, then we have a homomorphism $rho: (C^*)^n rightarrow GL(V)$ whose image consists entirely of semisimple elements. So now, we show that if an element $g in GL(V)$ is semisimple then it acts diagonally on $V$. This will then imply that each element of $rho((C^*)^n)$ acts diagonally on $V$ as and hence the entire action can be simultaneously diagonalized since the operators also commute with each other.
To finish the proof, we use the multiplicative Jordan decomposition. This can be found on page 96 of Humphreys (and many other sources I would imagine.) This says that for any element $g in GL(V)$, there exist $g_s, g_u in GL(V)$ such that
- $g, g_s, g_u$ commute with any linear map $Vrightarrow V$ that commutes with $g$.
- $g_s$ is diagonalizable.
- $g_u$ is unipotent i.e. there exists some $m$ such that $(g_u - 1)^m = 0.$
So now, since $g_s$ is diagonalizable, it must act semisimply on $C[GL(V)]$. This is because we let $x_i, j$ be coordinates for $GL(V) cong M(n times n)$ in the basis such that $g_s$ is diagonal, then $C[GL(V)] = C[X_1,1, ldots, x_n, n]_det$ and the elements of the form
$$fracx_1, 1^a_1, 1cdots x_n, n^a_n, ndet^l$$
form a spanning set in $C[GL(V)]$ of eigenvectors for $g_s$ and hence for $g_s^-1.$ As $g$ also acts diagonally on $C[GL(V)]$, and commutes with $g_s^-1$, the action of $g$ and $g_s^-1$ on $C[GL(V)]$ must be simulataneously diagonalizable. Hence, $gg_s^-1 = g_u$ must also act diagonally on $C[GL(V)]$.
But $g_u - 1$ acts nilpotently on $C[GL(V)]$ because its $m$th power is $0$. Hence, the action of $g_u$ in $C[GL(V)]$ must have eigenvalue $1$ and is hence the identity. Since the representation of an algebraic group in its coordinate ring is faithful, this implies that $g_u = 1$ and hence $g = g_s$ is diagonalizable in $GL(V)$. This is what we needed.
answered Jun 10 '14 at 7:43
Siddharth Venkatesh
1,97565
answered Jun 10 '14 at 7:43
Siddharth Venkatesh
1,97565
answered Jun 10 '14 at 7:43
Siddharth Venkatesh
1,97565
answered Jun 10 '14 at 7:43
Siddharth Venkatesh
1,97565
1,97565
This is useful, thank you. I had one question. I know that a finite set of commuting diagonalizable linear operators can be simultaneously diagonalized. But $(mathbbC^ast)^n$ has infinitely many elements, so what is the argument that they can all be simultaneously diagonilized?
– ykm
Jun 11 '14 at 4:57
$(mathbbC^*)^n$ is the Zariski closure of the subgroup generated by finitely many elements. If these finitely many elements are diagonalized, so is the subgroup they generate and hence so is its Zariski closure as the set of diagonal matrices is Zariski closed.
– Siddharth Venkatesh
Jun 11 '14 at 8:08
add a comment |Â
This is useful, thank you. I had one question. I know that a finite set of commuting diagonalizable linear operators can be simultaneously diagonalized. But $(mathbbC^ast)^n$ has infinitely many elements, so what is the argument that they can all be simultaneously diagonilized?
– ykm
Jun 11 '14 at 4:57
$(mathbbC^*)^n$ is the Zariski closure of the subgroup generated by finitely many elements. If these finitely many elements are diagonalized, so is the subgroup they generate and hence so is its Zariski closure as the set of diagonal matrices is Zariski closed.
– Siddharth Venkatesh
Jun 11 '14 at 8:08
This is useful, thank you. I had one question. I know that a finite set of commuting diagonalizable linear operators can be simultaneously diagonalized. But $(mathbbC^ast)^n$ has infinitely many elements, so what is the argument that they can all be simultaneously diagonilized?
– ykm
Jun 11 '14 at 4:57
This is useful, thank you. I had one question. I know that a finite set of commuting diagonalizable linear operators can be simultaneously diagonalized. But $(mathbbC^ast)^n$ has infinitely many elements, so what is the argument that they can all be simultaneously diagonilized?
– ykm
Jun 11 '14 at 4:57
$(mathbbC^*)^n$ is the Zariski closure of the subgroup generated by finitely many elements. If these finitely many elements are diagonalized, so is the subgroup they generate and hence so is its Zariski closure as the set of diagonal matrices is Zariski closed.
– Siddharth Venkatesh
Jun 11 '14 at 8:08
$(mathbbC^*)^n$ is the Zariski closure of the subgroup generated by finitely many elements. If these finitely many elements are diagonalized, so is the subgroup they generate and hence so is its Zariski closure as the set of diagonal matrices is Zariski closed.
– Siddharth Venkatesh
Jun 11 '14 at 8:08
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