Dimension of the invariant subspaceDimension of Hom(U, V)Invariant polynomials over symmetric matrices under Euclidean transformationsIs the ring of polynomial invariants of a finite perfect group an UFD?Invariants of finite groupsDoes the invariant ring determine the group?Reference request: invariant theoryInvariant polynomials under an $S_4$-actionFind irreducible representations of semidirect product $(S_2 times S_2) rtimes S_2$Explicitly finding a quotient of two polynomial spacesOn the Definition of the Reynolds Operator.
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Dimension of the invariant subspace
Dimension of Hom(U, V)Invariant polynomials over symmetric matrices under Euclidean transformationsIs the ring of polynomial invariants of a finite perfect group an UFD?Invariants of finite groupsDoes the invariant ring determine the group?Reference request: invariant theoryInvariant polynomials under an $S_4$-actionFind irreducible representations of semidirect product $(S_2 times S_2) rtimes S_2$Explicitly finding a quotient of two polynomial spacesOn the Definition of the Reynolds Operator.
$begingroup$
Let $Gamma subseteq GL_n(mathbbC)$ be a finite matrix group. Let this finite matrix group act on $f(x_1,...,x_n) in mathbbC[x_1,...,x_n]$ like so: $$Gamma cdot f(x_1,...,x_n) = f(Gamma textbfx)$$ where $textbfx$ is to be thought of as the column vector of the variables $x_1,...,x_n$.
Define the invariant subspace $mathbbC[x_1,...,x_n]^Gamma = f in mathbbC[x_1,...,x_n] : A cdot f = f hspace2mm forall hspace2mm A in Gamma$.
Now, define the Reynold's operator $R_Gamma : mathbbC[x_1,...,x_n] rightarrow mathbbC[x_1,...,x_n]$ by: $$R_Gamma (f)(textbfx) = frac1 sum_A in Gamma f(A textbfx)$$
Now, the number of linearly independent invariants of $Gamma$ of degree $1$ is given by $$a_1 = frac1 sum_A in Gamma trace(A)$$ But I'm not sure why this is so? I know that $R_Gamma$ is projection on to $mathbbC[x_1,...,x_n]$ and $im(R_Gamma) = mathbbC[x_1,...,x_n]^Gamma$, and so this would imply that $trace(R_Gamma) = dim(mathbbC[x_1,...,x_n]^Gamma)$, but where do I go from here? What is the trace of this Reynold's Operator?
I'm not even sure if I'm going in the right direction here, because I'm not sure why this $dim(mathbbC[x_1,...,x_n]^Gamma)$ would even give the number of linearly independent invariants of $Gamma$ of degree $1$. Where does the degree $1$ bit come from?
abstract-algebra representation-theory invariant-theory invariance
asked Mar 29 at 21:20
the manthe man
831716
$endgroup$
|
show 5 more comments
$begingroup$
Let $Gamma subseteq GL_n(mathbbC)$ be a finite matrix group. Let this finite matrix group act on $f(x_1,...,x_n) in mathbbC[x_1,...,x_n]$ like so: $$Gamma cdot f(x_1,...,x_n) = f(Gamma textbfx)$$ where $textbfx$ is to be thought of as the column vector of the variables $x_1,...,x_n$.
Define the invariant subspace $mathbbC[x_1,...,x_n]^Gamma = f in mathbbC[x_1,...,x_n] : A cdot f = f hspace2mm forall hspace2mm A in Gamma$.
Now, define the Reynold's operator $R_Gamma : mathbbC[x_1,...,x_n] rightarrow mathbbC[x_1,...,x_n]$ by: $$R_Gamma (f)(textbfx) = frac1 sum_A in Gamma f(A textbfx)$$
Now, the number of linearly independent invariants of $Gamma$ of degree $1$ is given by $$a_1 = frac1 sum_A in Gamma trace(A)$$ But I'm not sure why this is so? I know that $R_Gamma$ is projection on to $mathbbC[x_1,...,x_n]$ and $im(R_Gamma) = mathbbC[x_1,...,x_n]^Gamma$, and so this would imply that $trace(R_Gamma) = dim(mathbbC[x_1,...,x_n]^Gamma)$, but where do I go from here? What is the trace of this Reynold's Operator?
I'm not even sure if I'm going in the right direction here, because I'm not sure why this $dim(mathbbC[x_1,...,x_n]^Gamma)$ would even give the number of linearly independent invariants of $Gamma$ of degree $1$. Where does the degree $1$ bit come from?
abstract-algebra representation-theory invariant-theory invariance
asked Mar 29 at 21:20
the manthe man
831716
$endgroup$
2
$begingroup$
The left action of a $gamma$ on $f$ is $gamma cdot f(textbfx )= f(gamma^-1 textbfx)$. Otherwise it is a right-action.
$endgroup$
– AlexL
Mar 29 at 23:12
1
$begingroup$
If $R_Gamma$ is a projection, then its restriction $r_Gamma : mathbbC_leqslant 1[textbfx] to mathbbC_leqslant 1[textbfx]$ (the polynomials of degree $leqslant 1$) is the projection on $mathbbC_leqslant 1[textbfx]^Gamma$ and $a_1=dim(mathbbC_leqslant 1[textbfx]^Gamma)=rk(r_Gamma)=trace(r_Gamma)$. To calcultate $trace(r_Gamma)$, use the basis $(x_1,cdots,x_n)$ of $mathbbC_leqslant 1[textbfx]$. It should lead you to the result
$endgroup$
– AlexL
Mar 29 at 23:12
$begingroup$
How do you mean use the basis?
$endgroup$
– the man
Mar 29 at 23:49
$begingroup$
Also, in all the literature I’ve read, when discussing invariance, they let a matrix $M$ act on a polynomial $f(x_1,...,x_n)$ as $M cdot f(x_1,...,x_n) = f(M textbfx)$?
$endgroup$
– the man
Mar 29 at 23:57
$begingroup$
@theman Try computing $MNcdot f$ and $Mcdot (Ncdot f)$ with that definition. You'll notice that $(MNcdot f)x=f(MNx)$, whereas $(Mcdot (Ncdot f))x=(Ncdot f)(Mx)=f(NMx)$. These are not equal.
$endgroup$
– jgon
Mar 30 at 0:04
|
show 5 more comments
$begingroup$
Let $Gamma subseteq GL_n(mathbbC)$ be a finite matrix group. Let this finite matrix group act on $f(x_1,...,x_n) in mathbbC[x_1,...,x_n]$ like so: $$Gamma cdot f(x_1,...,x_n) = f(Gamma textbfx)$$ where $textbfx$ is to be thought of as the column vector of the variables $x_1,...,x_n$.
Define the invariant subspace $mathbbC[x_1,...,x_n]^Gamma = f in mathbbC[x_1,...,x_n] : A cdot f = f hspace2mm forall hspace2mm A in Gamma$.
Now, define the Reynold's operator $R_Gamma : mathbbC[x_1,...,x_n] rightarrow mathbbC[x_1,...,x_n]$ by: $$R_Gamma (f)(textbfx) = frac1 sum_A in Gamma f(A textbfx)$$
Now, the number of linearly independent invariants of $Gamma$ of degree $1$ is given by $$a_1 = frac1 sum_A in Gamma trace(A)$$ But I'm not sure why this is so? I know that $R_Gamma$ is projection on to $mathbbC[x_1,...,x_n]$ and $im(R_Gamma) = mathbbC[x_1,...,x_n]^Gamma$, and so this would imply that $trace(R_Gamma) = dim(mathbbC[x_1,...,x_n]^Gamma)$, but where do I go from here? What is the trace of this Reynold's Operator?
I'm not even sure if I'm going in the right direction here, because I'm not sure why this $dim(mathbbC[x_1,...,x_n]^Gamma)$ would even give the number of linearly independent invariants of $Gamma$ of degree $1$. Where does the degree $1$ bit come from?
abstract-algebra representation-theory invariant-theory invariance
asked Mar 29 at 21:20
the manthe man
831716
$endgroup$
Let $Gamma subseteq GL_n(mathbbC)$ be a finite matrix group. Let this finite matrix group act on $f(x_1,...,x_n) in mathbbC[x_1,...,x_n]$ like so: $$Gamma cdot f(x_1,...,x_n) = f(Gamma textbfx)$$ where $textbfx$ is to be thought of as the column vector of the variables $x_1,...,x_n$.
Define the invariant subspace $mathbbC[x_1,...,x_n]^Gamma = f in mathbbC[x_1,...,x_n] : A cdot f = f hspace2mm forall hspace2mm A in Gamma$.
Now, define the Reynold's operator $R_Gamma : mathbbC[x_1,...,x_n] rightarrow mathbbC[x_1,...,x_n]$ by: $$R_Gamma (f)(textbfx) = frac1 sum_A in Gamma f(A textbfx)$$
Now, the number of linearly independent invariants of $Gamma$ of degree $1$ is given by $$a_1 = frac1 sum_A in Gamma trace(A)$$ But I'm not sure why this is so? I know that $R_Gamma$ is projection on to $mathbbC[x_1,...,x_n]$ and $im(R_Gamma) = mathbbC[x_1,...,x_n]^Gamma$, and so this would imply that $trace(R_Gamma) = dim(mathbbC[x_1,...,x_n]^Gamma)$, but where do I go from here? What is the trace of this Reynold's Operator?
I'm not even sure if I'm going in the right direction here, because I'm not sure why this $dim(mathbbC[x_1,...,x_n]^Gamma)$ would even give the number of linearly independent invariants of $Gamma$ of degree $1$. Where does the degree $1$ bit come from?
abstract-algebra representation-theory invariant-theory invariance
abstract-algebra representation-theory invariant-theory invariance
asked Mar 29 at 21:20
the manthe man
831716
asked Mar 29 at 21:20
the manthe man
831716
edited Mar 30 at 0:16
the man
asked Mar 29 at 21:20
the manthe man
831716
asked Mar 29 at 21:20
the manthe man
831716
asked Mar 29 at 21:20
the manthe man
831716
831716
2
$begingroup$
The left action of a $gamma$ on $f$ is $gamma cdot f(textbfx )= f(gamma^-1 textbfx)$. Otherwise it is a right-action.
$endgroup$
– AlexL
Mar 29 at 23:12
1
$begingroup$
If $R_Gamma$ is a projection, then its restriction $r_Gamma : mathbbC_leqslant 1[textbfx] to mathbbC_leqslant 1[textbfx]$ (the polynomials of degree $leqslant 1$) is the projection on $mathbbC_leqslant 1[textbfx]^Gamma$ and $a_1=dim(mathbbC_leqslant 1[textbfx]^Gamma)=rk(r_Gamma)=trace(r_Gamma)$. To calcultate $trace(r_Gamma)$, use the basis $(x_1,cdots,x_n)$ of $mathbbC_leqslant 1[textbfx]$. It should lead you to the result
$endgroup$
– AlexL
Mar 29 at 23:12
$begingroup$
How do you mean use the basis?
$endgroup$
– the man
Mar 29 at 23:49
$begingroup$
Also, in all the literature I’ve read, when discussing invariance, they let a matrix $M$ act on a polynomial $f(x_1,...,x_n)$ as $M cdot f(x_1,...,x_n) = f(M textbfx)$?
$endgroup$
– the man
Mar 29 at 23:57
$begingroup$
@theman Try computing $MNcdot f$ and $Mcdot (Ncdot f)$ with that definition. You'll notice that $(MNcdot f)x=f(MNx)$, whereas $(Mcdot (Ncdot f))x=(Ncdot f)(Mx)=f(NMx)$. These are not equal.
$endgroup$
– jgon
Mar 30 at 0:04
|
show 5 more comments
2
$begingroup$
The left action of a $gamma$ on $f$ is $gamma cdot f(textbfx )= f(gamma^-1 textbfx)$. Otherwise it is a right-action.
$endgroup$
– AlexL
Mar 29 at 23:12
1
$begingroup$
If $R_Gamma$ is a projection, then its restriction $r_Gamma : mathbbC_leqslant 1[textbfx] to mathbbC_leqslant 1[textbfx]$ (the polynomials of degree $leqslant 1$) is the projection on $mathbbC_leqslant 1[textbfx]^Gamma$ and $a_1=dim(mathbbC_leqslant 1[textbfx]^Gamma)=rk(r_Gamma)=trace(r_Gamma)$. To calcultate $trace(r_Gamma)$, use the basis $(x_1,cdots,x_n)$ of $mathbbC_leqslant 1[textbfx]$. It should lead you to the result
$endgroup$
– AlexL
Mar 29 at 23:12
$begingroup$
How do you mean use the basis?
$endgroup$
– the man
Mar 29 at 23:49
$begingroup$
Also, in all the literature I’ve read, when discussing invariance, they let a matrix $M$ act on a polynomial $f(x_1,...,x_n)$ as $M cdot f(x_1,...,x_n) = f(M textbfx)$?
$endgroup$
– the man
Mar 29 at 23:57
$begingroup$
@theman Try computing $MNcdot f$ and $Mcdot (Ncdot f)$ with that definition. You'll notice that $(MNcdot f)x=f(MNx)$, whereas $(Mcdot (Ncdot f))x=(Ncdot f)(Mx)=f(NMx)$. These are not equal.
$endgroup$
– jgon
Mar 30 at 0:04
2
2
$begingroup$
The left action of a $gamma$ on $f$ is $gamma cdot f(textbfx )= f(gamma^-1 textbfx)$. Otherwise it is a right-action.
$endgroup$
– AlexL
Mar 29 at 23:12
$begingroup$
The left action of a $gamma$ on $f$ is $gamma cdot f(textbfx )= f(gamma^-1 textbfx)$. Otherwise it is a right-action.
$endgroup$
– AlexL
Mar 29 at 23:12
1
1
$begingroup$
If $R_Gamma$ is a projection, then its restriction $r_Gamma : mathbbC_leqslant 1[textbfx] to mathbbC_leqslant 1[textbfx]$ (the polynomials of degree $leqslant 1$) is the projection on $mathbbC_leqslant 1[textbfx]^Gamma$ and $a_1=dim(mathbbC_leqslant 1[textbfx]^Gamma)=rk(r_Gamma)=trace(r_Gamma)$. To calcultate $trace(r_Gamma)$, use the basis $(x_1,cdots,x_n)$ of $mathbbC_leqslant 1[textbfx]$. It should lead you to the result
$endgroup$
– AlexL
Mar 29 at 23:12
$begingroup$
If $R_Gamma$ is a projection, then its restriction $r_Gamma : mathbbC_leqslant 1[textbfx] to mathbbC_leqslant 1[textbfx]$ (the polynomials of degree $leqslant 1$) is the projection on $mathbbC_leqslant 1[textbfx]^Gamma$ and $a_1=dim(mathbbC_leqslant 1[textbfx]^Gamma)=rk(r_Gamma)=trace(r_Gamma)$. To calcultate $trace(r_Gamma)$, use the basis $(x_1,cdots,x_n)$ of $mathbbC_leqslant 1[textbfx]$. It should lead you to the result
$endgroup$
– AlexL
Mar 29 at 23:12
$begingroup$
How do you mean use the basis?
$endgroup$
– the man
Mar 29 at 23:49
$begingroup$
How do you mean use the basis?
$endgroup$
– the man
Mar 29 at 23:49
$begingroup$
Also, in all the literature I’ve read, when discussing invariance, they let a matrix $M$ act on a polynomial $f(x_1,...,x_n)$ as $M cdot f(x_1,...,x_n) = f(M textbfx)$?
$endgroup$
– the man
Mar 29 at 23:57
$begingroup$
Also, in all the literature I’ve read, when discussing invariance, they let a matrix $M$ act on a polynomial $f(x_1,...,x_n)$ as $M cdot f(x_1,...,x_n) = f(M textbfx)$?
$endgroup$
– the man
Mar 29 at 23:57
$begingroup$
@theman Try computing $MNcdot f$ and $Mcdot (Ncdot f)$ with that definition. You'll notice that $(MNcdot f)x=f(MNx)$, whereas $(Mcdot (Ncdot f))x=(Ncdot f)(Mx)=f(NMx)$. These are not equal.
$endgroup$
– jgon
Mar 30 at 0:04
$begingroup$
@theman Try computing $MNcdot f$ and $Mcdot (Ncdot f)$ with that definition. You'll notice that $(MNcdot f)x=f(MNx)$, whereas $(Mcdot (Ncdot f))x=(Ncdot f)(Mx)=f(NMx)$. These are not equal.
$endgroup$
– jgon
Mar 30 at 0:04
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
In those cases I need elementary discussions
For any representation of finite group $rho : G to GL(V)$ to inversible linear maps of a $BbbC$-vector space, then $P=frac1sum_g in Grho(g)$ is a projection of $V$ on the $G$-fixed subspace $V^G$ (proof : if $v in V$ then $Pv in V^G$ and if $v in V^G$ then $Pv=v$)
in some basis $B$ you'll have $P = B pmatrixI_m & 0 \ 0 & 0 B^-1$ where $m = dim V^G$ so $trace(P) = trace( pmatrixI_m & 0 \ 0 & 0) = dim V^G$.
You need to make clear you are considering $V =BbbC^n$ and the corresponding $trace$, no polynomial ring.
From there you can construct other representations on $BbbC[x_1,ldots,x_n]_d$ the set of homogeneous polynomials of degree $d$, the obtained representation $pi(g)(f(x))= f(rho(g)x)$ is called $pi = Sym^drho$, and what you defined is the natural infinite dimensional rep. $bigoplus_d Sym^drho$ of $G=Gamma$ on $BbbC[x_1,ldots,x_n] = bigoplus_d BbbC[x_1,ldots,x_n]_d$.
Then the point is that $V = V^G oplus W$ where $W = ker(P)$ and $W$ is sent to itself by the $Ain Gamma$ thus is a subrepresentation. This decomposition translates to the polynomials obtaining that with the linear polynomials $(y_1,ldots,y_m,z_1,ldots,z_n-m) = B(x_1,ldots,x_n)$ : $BbbC[x_1,ldots,x_n]= BbbC[y_1,ldots,y_m,z_1,ldots,z_n-m]$ and $A.f(y_1,ldots,y_m,z_1,ldots,z_n-m) = f((y_1,ldots,y_m,0,ldots)+BA B^-1 (0,ldots,z_1,ldots,z_n-m))$.
If $G$ is a finite group then $BbbC[x_1,ldots,x_n]/BbbC[x_1,ldots,x_n]^G$ is a finite Galois extension with Galois group $H=G/ker(rho)$ so $BbbC[x_1,ldots,x_n]^G=BbbC[y_1,ldots,y_m,f_1,ldots,f_n-m]$ for some algebraically independent polynomials $f$ (of degree $> 1$). Not sure how to find $BbbC[x_1,ldots,x_n]^G$ and its transcendental degree when $H$ is infinite.
answered Mar 30 at 3:41
reunsreuns
20.6k21352
$endgroup$
$begingroup$
So, where I've written $trace(R_Gamma) = dim(mathbbC[x_1,...,x_n]^G)$, does this not maky any sense?
$endgroup$
– the man
Mar 30 at 10:42
1
$begingroup$
No, except if you meant the Krull dimension of the ring $mathbbC[x_1,...,x_n]^G$ (the size of its transcendental basis over $BbbC$) in which case yes, which is what I did with $mathbbC[x_1,...,x_n]^G = BbbC[y_1,ldots,y_m]$
$endgroup$
– reuns
Mar 30 at 12:47
$begingroup$
Why does it not make any sense? Can we not think of it as a vector space? I mean, the quantity I'm after is $a_1 = dim_mathbbC(mathbbC[x_1,...,x_n]_1^Gamma)$ right?
$endgroup$
– the man
Mar 30 at 13:00
1
$begingroup$
@theman $mathbbC[x_1,...,x_n]^G$ is a ring of polynomials, it is an infinite dimensional vector space. By the way if $G$ is a finite group then $mathbbC[x_1,...,x_n]^G$ always contains more than $BbbC[y_1,ldots,y_m]$, $mathbbC[x_1,...,x_n]/mathbbC[x_1,...,x_n]^G$ is a Galois extension of degree $|G/ker(rho)|$ thus the transcendental degree of $mathbbC[x_1,...,x_n]^G$ is $n$, not $m$ (concretely for some $f_1,ldots,f_G/ker(rho)$ then $mathbbC[x_1,...,x_n]=sum_j=1^G/ker(rho) mathbbC[x_1,...,x_n]^G f_j(x)$)
$endgroup$
– reuns
Mar 30 at 13:08
$begingroup$
@ Oh yes, I see. Thank you very much!
$endgroup$
– the man
Mar 30 at 13:10
|
show 1 more comment
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1 Answer
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$begingroup$
In those cases I need elementary discussions
For any representation of finite group $rho : G to GL(V)$ to inversible linear maps of a $BbbC$-vector space, then $P=frac1sum_g in Grho(g)$ is a projection of $V$ on the $G$-fixed subspace $V^G$ (proof : if $v in V$ then $Pv in V^G$ and if $v in V^G$ then $Pv=v$)
in some basis $B$ you'll have $P = B pmatrixI_m & 0 \ 0 & 0 B^-1$ where $m = dim V^G$ so $trace(P) = trace( pmatrixI_m & 0 \ 0 & 0) = dim V^G$.
You need to make clear you are considering $V =BbbC^n$ and the corresponding $trace$, no polynomial ring.
From there you can construct other representations on $BbbC[x_1,ldots,x_n]_d$ the set of homogeneous polynomials of degree $d$, the obtained representation $pi(g)(f(x))= f(rho(g)x)$ is called $pi = Sym^drho$, and what you defined is the natural infinite dimensional rep. $bigoplus_d Sym^drho$ of $G=Gamma$ on $BbbC[x_1,ldots,x_n] = bigoplus_d BbbC[x_1,ldots,x_n]_d$.
Then the point is that $V = V^G oplus W$ where $W = ker(P)$ and $W$ is sent to itself by the $Ain Gamma$ thus is a subrepresentation. This decomposition translates to the polynomials obtaining that with the linear polynomials $(y_1,ldots,y_m,z_1,ldots,z_n-m) = B(x_1,ldots,x_n)$ : $BbbC[x_1,ldots,x_n]= BbbC[y_1,ldots,y_m,z_1,ldots,z_n-m]$ and $A.f(y_1,ldots,y_m,z_1,ldots,z_n-m) = f((y_1,ldots,y_m,0,ldots)+BA B^-1 (0,ldots,z_1,ldots,z_n-m))$.
If $G$ is a finite group then $BbbC[x_1,ldots,x_n]/BbbC[x_1,ldots,x_n]^G$ is a finite Galois extension with Galois group $H=G/ker(rho)$ so $BbbC[x_1,ldots,x_n]^G=BbbC[y_1,ldots,y_m,f_1,ldots,f_n-m]$ for some algebraically independent polynomials $f$ (of degree $> 1$). Not sure how to find $BbbC[x_1,ldots,x_n]^G$ and its transcendental degree when $H$ is infinite.
answered Mar 30 at 3:41
reunsreuns
20.6k21352
$endgroup$
$begingroup$
So, where I've written $trace(R_Gamma) = dim(mathbbC[x_1,...,x_n]^G)$, does this not maky any sense?
$endgroup$
– the man
Mar 30 at 10:42
1
$begingroup$
No, except if you meant the Krull dimension of the ring $mathbbC[x_1,...,x_n]^G$ (the size of its transcendental basis over $BbbC$) in which case yes, which is what I did with $mathbbC[x_1,...,x_n]^G = BbbC[y_1,ldots,y_m]$
$endgroup$
– reuns
Mar 30 at 12:47
$begingroup$
Why does it not make any sense? Can we not think of it as a vector space? I mean, the quantity I'm after is $a_1 = dim_mathbbC(mathbbC[x_1,...,x_n]_1^Gamma)$ right?
$endgroup$
– the man
Mar 30 at 13:00
1
$begingroup$
@theman $mathbbC[x_1,...,x_n]^G$ is a ring of polynomials, it is an infinite dimensional vector space. By the way if $G$ is a finite group then $mathbbC[x_1,...,x_n]^G$ always contains more than $BbbC[y_1,ldots,y_m]$, $mathbbC[x_1,...,x_n]/mathbbC[x_1,...,x_n]^G$ is a Galois extension of degree $|G/ker(rho)|$ thus the transcendental degree of $mathbbC[x_1,...,x_n]^G$ is $n$, not $m$ (concretely for some $f_1,ldots,f_G/ker(rho)$ then $mathbbC[x_1,...,x_n]=sum_j=1^G/ker(rho) mathbbC[x_1,...,x_n]^G f_j(x)$)
$endgroup$
– reuns
Mar 30 at 13:08
$begingroup$
@ Oh yes, I see. Thank you very much!
$endgroup$
– the man
Mar 30 at 13:10
|
show 1 more comment
$begingroup$
In those cases I need elementary discussions
For any representation of finite group $rho : G to GL(V)$ to inversible linear maps of a $BbbC$-vector space, then $P=frac1sum_g in Grho(g)$ is a projection of $V$ on the $G$-fixed subspace $V^G$ (proof : if $v in V$ then $Pv in V^G$ and if $v in V^G$ then $Pv=v$)
in some basis $B$ you'll have $P = B pmatrixI_m & 0 \ 0 & 0 B^-1$ where $m = dim V^G$ so $trace(P) = trace( pmatrixI_m & 0 \ 0 & 0) = dim V^G$.
You need to make clear you are considering $V =BbbC^n$ and the corresponding $trace$, no polynomial ring.
From there you can construct other representations on $BbbC[x_1,ldots,x_n]_d$ the set of homogeneous polynomials of degree $d$, the obtained representation $pi(g)(f(x))= f(rho(g)x)$ is called $pi = Sym^drho$, and what you defined is the natural infinite dimensional rep. $bigoplus_d Sym^drho$ of $G=Gamma$ on $BbbC[x_1,ldots,x_n] = bigoplus_d BbbC[x_1,ldots,x_n]_d$.
Then the point is that $V = V^G oplus W$ where $W = ker(P)$ and $W$ is sent to itself by the $Ain Gamma$ thus is a subrepresentation. This decomposition translates to the polynomials obtaining that with the linear polynomials $(y_1,ldots,y_m,z_1,ldots,z_n-m) = B(x_1,ldots,x_n)$ : $BbbC[x_1,ldots,x_n]= BbbC[y_1,ldots,y_m,z_1,ldots,z_n-m]$ and $A.f(y_1,ldots,y_m,z_1,ldots,z_n-m) = f((y_1,ldots,y_m,0,ldots)+BA B^-1 (0,ldots,z_1,ldots,z_n-m))$.
If $G$ is a finite group then $BbbC[x_1,ldots,x_n]/BbbC[x_1,ldots,x_n]^G$ is a finite Galois extension with Galois group $H=G/ker(rho)$ so $BbbC[x_1,ldots,x_n]^G=BbbC[y_1,ldots,y_m,f_1,ldots,f_n-m]$ for some algebraically independent polynomials $f$ (of degree $> 1$). Not sure how to find $BbbC[x_1,ldots,x_n]^G$ and its transcendental degree when $H$ is infinite.
answered Mar 30 at 3:41
reunsreuns
20.6k21352
$endgroup$
$begingroup$
So, where I've written $trace(R_Gamma) = dim(mathbbC[x_1,...,x_n]^G)$, does this not maky any sense?
$endgroup$
– the man
Mar 30 at 10:42
1
$begingroup$
No, except if you meant the Krull dimension of the ring $mathbbC[x_1,...,x_n]^G$ (the size of its transcendental basis over $BbbC$) in which case yes, which is what I did with $mathbbC[x_1,...,x_n]^G = BbbC[y_1,ldots,y_m]$
$endgroup$
– reuns
Mar 30 at 12:47
$begingroup$
Why does it not make any sense? Can we not think of it as a vector space? I mean, the quantity I'm after is $a_1 = dim_mathbbC(mathbbC[x_1,...,x_n]_1^Gamma)$ right?
$endgroup$
– the man
Mar 30 at 13:00
1
$begingroup$
@theman $mathbbC[x_1,...,x_n]^G$ is a ring of polynomials, it is an infinite dimensional vector space. By the way if $G$ is a finite group then $mathbbC[x_1,...,x_n]^G$ always contains more than $BbbC[y_1,ldots,y_m]$, $mathbbC[x_1,...,x_n]/mathbbC[x_1,...,x_n]^G$ is a Galois extension of degree $|G/ker(rho)|$ thus the transcendental degree of $mathbbC[x_1,...,x_n]^G$ is $n$, not $m$ (concretely for some $f_1,ldots,f_G/ker(rho)$ then $mathbbC[x_1,...,x_n]=sum_j=1^G/ker(rho) mathbbC[x_1,...,x_n]^G f_j(x)$)
$endgroup$
– reuns
Mar 30 at 13:08
$begingroup$
@ Oh yes, I see. Thank you very much!
$endgroup$
– the man
Mar 30 at 13:10
|
show 1 more comment
$begingroup$
In those cases I need elementary discussions
For any representation of finite group $rho : G to GL(V)$ to inversible linear maps of a $BbbC$-vector space, then $P=frac1sum_g in Grho(g)$ is a projection of $V$ on the $G$-fixed subspace $V^G$ (proof : if $v in V$ then $Pv in V^G$ and if $v in V^G$ then $Pv=v$)
in some basis $B$ you'll have $P = B pmatrixI_m & 0 \ 0 & 0 B^-1$ where $m = dim V^G$ so $trace(P) = trace( pmatrixI_m & 0 \ 0 & 0) = dim V^G$.
You need to make clear you are considering $V =BbbC^n$ and the corresponding $trace$, no polynomial ring.
From there you can construct other representations on $BbbC[x_1,ldots,x_n]_d$ the set of homogeneous polynomials of degree $d$, the obtained representation $pi(g)(f(x))= f(rho(g)x)$ is called $pi = Sym^drho$, and what you defined is the natural infinite dimensional rep. $bigoplus_d Sym^drho$ of $G=Gamma$ on $BbbC[x_1,ldots,x_n] = bigoplus_d BbbC[x_1,ldots,x_n]_d$.
Then the point is that $V = V^G oplus W$ where $W = ker(P)$ and $W$ is sent to itself by the $Ain Gamma$ thus is a subrepresentation. This decomposition translates to the polynomials obtaining that with the linear polynomials $(y_1,ldots,y_m,z_1,ldots,z_n-m) = B(x_1,ldots,x_n)$ : $BbbC[x_1,ldots,x_n]= BbbC[y_1,ldots,y_m,z_1,ldots,z_n-m]$ and $A.f(y_1,ldots,y_m,z_1,ldots,z_n-m) = f((y_1,ldots,y_m,0,ldots)+BA B^-1 (0,ldots,z_1,ldots,z_n-m))$.
If $G$ is a finite group then $BbbC[x_1,ldots,x_n]/BbbC[x_1,ldots,x_n]^G$ is a finite Galois extension with Galois group $H=G/ker(rho)$ so $BbbC[x_1,ldots,x_n]^G=BbbC[y_1,ldots,y_m,f_1,ldots,f_n-m]$ for some algebraically independent polynomials $f$ (of degree $> 1$). Not sure how to find $BbbC[x_1,ldots,x_n]^G$ and its transcendental degree when $H$ is infinite.
answered Mar 30 at 3:41
reunsreuns
20.6k21352
$endgroup$
In those cases I need elementary discussions
For any representation of finite group $rho : G to GL(V)$ to inversible linear maps of a $BbbC$-vector space, then $P=frac1sum_g in Grho(g)$ is a projection of $V$ on the $G$-fixed subspace $V^G$ (proof : if $v in V$ then $Pv in V^G$ and if $v in V^G$ then $Pv=v$)
in some basis $B$ you'll have $P = B pmatrixI_m & 0 \ 0 & 0 B^-1$ where $m = dim V^G$ so $trace(P) = trace( pmatrixI_m & 0 \ 0 & 0) = dim V^G$.
You need to make clear you are considering $V =BbbC^n$ and the corresponding $trace$, no polynomial ring.
From there you can construct other representations on $BbbC[x_1,ldots,x_n]_d$ the set of homogeneous polynomials of degree $d$, the obtained representation $pi(g)(f(x))= f(rho(g)x)$ is called $pi = Sym^drho$, and what you defined is the natural infinite dimensional rep. $bigoplus_d Sym^drho$ of $G=Gamma$ on $BbbC[x_1,ldots,x_n] = bigoplus_d BbbC[x_1,ldots,x_n]_d$.
Then the point is that $V = V^G oplus W$ where $W = ker(P)$ and $W$ is sent to itself by the $Ain Gamma$ thus is a subrepresentation. This decomposition translates to the polynomials obtaining that with the linear polynomials $(y_1,ldots,y_m,z_1,ldots,z_n-m) = B(x_1,ldots,x_n)$ : $BbbC[x_1,ldots,x_n]= BbbC[y_1,ldots,y_m,z_1,ldots,z_n-m]$ and $A.f(y_1,ldots,y_m,z_1,ldots,z_n-m) = f((y_1,ldots,y_m,0,ldots)+BA B^-1 (0,ldots,z_1,ldots,z_n-m))$.
If $G$ is a finite group then $BbbC[x_1,ldots,x_n]/BbbC[x_1,ldots,x_n]^G$ is a finite Galois extension with Galois group $H=G/ker(rho)$ so $BbbC[x_1,ldots,x_n]^G=BbbC[y_1,ldots,y_m,f_1,ldots,f_n-m]$ for some algebraically independent polynomials $f$ (of degree $> 1$). Not sure how to find $BbbC[x_1,ldots,x_n]^G$ and its transcendental degree when $H$ is infinite.
answered Mar 30 at 3:41
reunsreuns
20.6k21352
edited Mar 30 at 13:13
answered Mar 30 at 3:41
reunsreuns
20.6k21352
answered Mar 30 at 3:41
reunsreuns
20.6k21352
answered Mar 30 at 3:41
reunsreuns
20.6k21352
20.6k21352
$begingroup$
So, where I've written $trace(R_Gamma) = dim(mathbbC[x_1,...,x_n]^G)$, does this not maky any sense?
$endgroup$
– the man
Mar 30 at 10:42
1
$begingroup$
No, except if you meant the Krull dimension of the ring $mathbbC[x_1,...,x_n]^G$ (the size of its transcendental basis over $BbbC$) in which case yes, which is what I did with $mathbbC[x_1,...,x_n]^G = BbbC[y_1,ldots,y_m]$
$endgroup$
– reuns
Mar 30 at 12:47
$begingroup$
Why does it not make any sense? Can we not think of it as a vector space? I mean, the quantity I'm after is $a_1 = dim_mathbbC(mathbbC[x_1,...,x_n]_1^Gamma)$ right?
$endgroup$
– the man
Mar 30 at 13:00
1
$begingroup$
@theman $mathbbC[x_1,...,x_n]^G$ is a ring of polynomials, it is an infinite dimensional vector space. By the way if $G$ is a finite group then $mathbbC[x_1,...,x_n]^G$ always contains more than $BbbC[y_1,ldots,y_m]$, $mathbbC[x_1,...,x_n]/mathbbC[x_1,...,x_n]^G$ is a Galois extension of degree $|G/ker(rho)|$ thus the transcendental degree of $mathbbC[x_1,...,x_n]^G$ is $n$, not $m$ (concretely for some $f_1,ldots,f_G/ker(rho)$ then $mathbbC[x_1,...,x_n]=sum_j=1^G/ker(rho) mathbbC[x_1,...,x_n]^G f_j(x)$)
$endgroup$
– reuns
Mar 30 at 13:08
$begingroup$
@ Oh yes, I see. Thank you very much!
$endgroup$
– the man
Mar 30 at 13:10
|
show 1 more comment
$begingroup$
So, where I've written $trace(R_Gamma) = dim(mathbbC[x_1,...,x_n]^G)$, does this not maky any sense?
$endgroup$
– the man
Mar 30 at 10:42
1
$begingroup$
No, except if you meant the Krull dimension of the ring $mathbbC[x_1,...,x_n]^G$ (the size of its transcendental basis over $BbbC$) in which case yes, which is what I did with $mathbbC[x_1,...,x_n]^G = BbbC[y_1,ldots,y_m]$
$endgroup$
– reuns
Mar 30 at 12:47
$begingroup$
Why does it not make any sense? Can we not think of it as a vector space? I mean, the quantity I'm after is $a_1 = dim_mathbbC(mathbbC[x_1,...,x_n]_1^Gamma)$ right?
$endgroup$
– the man
Mar 30 at 13:00
1
$begingroup$
@theman $mathbbC[x_1,...,x_n]^G$ is a ring of polynomials, it is an infinite dimensional vector space. By the way if $G$ is a finite group then $mathbbC[x_1,...,x_n]^G$ always contains more than $BbbC[y_1,ldots,y_m]$, $mathbbC[x_1,...,x_n]/mathbbC[x_1,...,x_n]^G$ is a Galois extension of degree $|G/ker(rho)|$ thus the transcendental degree of $mathbbC[x_1,...,x_n]^G$ is $n$, not $m$ (concretely for some $f_1,ldots,f_G/ker(rho)$ then $mathbbC[x_1,...,x_n]=sum_j=1^G/ker(rho) mathbbC[x_1,...,x_n]^G f_j(x)$)
$endgroup$
– reuns
Mar 30 at 13:08
$begingroup$
@ Oh yes, I see. Thank you very much!
$endgroup$
– the man
Mar 30 at 13:10
$begingroup$
So, where I've written $trace(R_Gamma) = dim(mathbbC[x_1,...,x_n]^G)$, does this not maky any sense?
$endgroup$
– the man
Mar 30 at 10:42
$begingroup$
So, where I've written $trace(R_Gamma) = dim(mathbbC[x_1,...,x_n]^G)$, does this not maky any sense?
$endgroup$
– the man
Mar 30 at 10:42
1
1
$begingroup$
No, except if you meant the Krull dimension of the ring $mathbbC[x_1,...,x_n]^G$ (the size of its transcendental basis over $BbbC$) in which case yes, which is what I did with $mathbbC[x_1,...,x_n]^G = BbbC[y_1,ldots,y_m]$
$endgroup$
– reuns
Mar 30 at 12:47
$begingroup$
No, except if you meant the Krull dimension of the ring $mathbbC[x_1,...,x_n]^G$ (the size of its transcendental basis over $BbbC$) in which case yes, which is what I did with $mathbbC[x_1,...,x_n]^G = BbbC[y_1,ldots,y_m]$
$endgroup$
– reuns
Mar 30 at 12:47
$begingroup$
Why does it not make any sense? Can we not think of it as a vector space? I mean, the quantity I'm after is $a_1 = dim_mathbbC(mathbbC[x_1,...,x_n]_1^Gamma)$ right?
$endgroup$
– the man
Mar 30 at 13:00
$begingroup$
Why does it not make any sense? Can we not think of it as a vector space? I mean, the quantity I'm after is $a_1 = dim_mathbbC(mathbbC[x_1,...,x_n]_1^Gamma)$ right?
$endgroup$
– the man
Mar 30 at 13:00
1
1
$begingroup$
@theman $mathbbC[x_1,...,x_n]^G$ is a ring of polynomials, it is an infinite dimensional vector space. By the way if $G$ is a finite group then $mathbbC[x_1,...,x_n]^G$ always contains more than $BbbC[y_1,ldots,y_m]$, $mathbbC[x_1,...,x_n]/mathbbC[x_1,...,x_n]^G$ is a Galois extension of degree $|G/ker(rho)|$ thus the transcendental degree of $mathbbC[x_1,...,x_n]^G$ is $n$, not $m$ (concretely for some $f_1,ldots,f_G/ker(rho)$ then $mathbbC[x_1,...,x_n]=sum_j=1^G/ker(rho) mathbbC[x_1,...,x_n]^G f_j(x)$)
$endgroup$
– reuns
Mar 30 at 13:08
$begingroup$
@theman $mathbbC[x_1,...,x_n]^G$ is a ring of polynomials, it is an infinite dimensional vector space. By the way if $G$ is a finite group then $mathbbC[x_1,...,x_n]^G$ always contains more than $BbbC[y_1,ldots,y_m]$, $mathbbC[x_1,...,x_n]/mathbbC[x_1,...,x_n]^G$ is a Galois extension of degree $|G/ker(rho)|$ thus the transcendental degree of $mathbbC[x_1,...,x_n]^G$ is $n$, not $m$ (concretely for some $f_1,ldots,f_G/ker(rho)$ then $mathbbC[x_1,...,x_n]=sum_j=1^G/ker(rho) mathbbC[x_1,...,x_n]^G f_j(x)$)
$endgroup$
– reuns
Mar 30 at 13:08
$begingroup$
@ Oh yes, I see. Thank you very much!
$endgroup$
– the man
Mar 30 at 13:10
$begingroup$
@ Oh yes, I see. Thank you very much!
$endgroup$
– the man
Mar 30 at 13:10
|
show 1 more comment
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$begingroup$
The left action of a $gamma$ on $f$ is $gamma cdot f(textbfx )= f(gamma^-1 textbfx)$. Otherwise it is a right-action.
$endgroup$
– AlexL
Mar 29 at 23:12
1
$begingroup$
If $R_Gamma$ is a projection, then its restriction $r_Gamma : mathbbC_leqslant 1[textbfx] to mathbbC_leqslant 1[textbfx]$ (the polynomials of degree $leqslant 1$) is the projection on $mathbbC_leqslant 1[textbfx]^Gamma$ and $a_1=dim(mathbbC_leqslant 1[textbfx]^Gamma)=rk(r_Gamma)=trace(r_Gamma)$. To calcultate $trace(r_Gamma)$, use the basis $(x_1,cdots,x_n)$ of $mathbbC_leqslant 1[textbfx]$. It should lead you to the result
$endgroup$
– AlexL
Mar 29 at 23:12
$begingroup$
How do you mean use the basis?
$endgroup$
– the man
Mar 29 at 23:49
$begingroup$
Also, in all the literature I’ve read, when discussing invariance, they let a matrix $M$ act on a polynomial $f(x_1,...,x_n)$ as $M cdot f(x_1,...,x_n) = f(M textbfx)$?
$endgroup$
– the man
Mar 29 at 23:57
$begingroup$
@theman Try computing $MNcdot f$ and $Mcdot (Ncdot f)$ with that definition. You'll notice that $(MNcdot f)x=f(MNx)$, whereas $(Mcdot (Ncdot f))x=(Ncdot f)(Mx)=f(NMx)$. These are not equal.
$endgroup$
– jgon
Mar 30 at 0:04