Representation of $V$ and polynomial rings The 2019 Stack Overflow Developer Survey Results Are InIs the dual representation of an irreducible representation always irreducible?Is every linear representation of a group $G$ on $k[x_1,dots,x_n]$ a dual representation?Equivalence of induced representationQuick question about formal definition of a representationDual space isomorphism and the dual representationPrimitive action vs. irreducible representationSymmetric polynomials in quotient of polynomial rings.Right-action representation definition, equivalence with left action repEquivalent representation definitionsGroup representation preserving finitely many generators
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Representation of $V$ and polynomial rings
The 2019 Stack Overflow Developer Survey Results Are InIs the dual representation of an irreducible representation always irreducible?Is every linear representation of a group $G$ on $k[x_1,dots,x_n]$ a dual representation?Equivalence of induced representationQuick question about formal definition of a representationDual space isomorphism and the dual representationPrimitive action vs. irreducible representationSymmetric polynomials in quotient of polynomial rings.Right-action representation definition, equivalence with left action repEquivalent representation definitionsGroup representation preserving finitely many generators
$begingroup$
Let $V$ be a complex vector space. Let $G$ be a finite group and let $$rho: G rightarrow GL(V)$$ be a representation of $G$.
Let $V^*$ denote the dual space of $V$, and let $mathcalO(V)$ denote the algebra of functions $F: V rightarrow mathbbC$ generated by the elements of $V^*$. Elements of $mathcalO(V)$ are called regular functions.
The dual representation $rho^*: G rightarrow GL(V^*)$ is given by $$(rho^*(g)f)(v) = f(rho(g)^-1v), forall g in G, f in V^*, v in V$$
Define the ring of invariant functions to be $$mathcalO(V)^G = f in mathcalO(V) : gf = f hspace2mm forall g in G$$
where $gf$ is just shorthand for $rho^*(g)f$.
Now, when we restrict to $V^* subset mathcalO(V)$, where $V^*$ has basis $x_1,...,x_n$, the regular functions are polynomials in the $x_i$ and the action of $G$ is given by $$gp(x_1,...,x_n) = p(gx_1,...,gx_n)$$ for some polynomial $p$.
$(1)$ - I don't understand this last bit? What is the distinction between $V^*$ and $mathcalO(V)$? Is $V^*$ just the set of $mathbbC$-linear sums of $x_1,...,x_n$? There is no multiplication between the $x_i$ defined? And an element of $mathcalO(V)$ is of the form $p(x_1,...,x_n)$ for any polynomial $p$ with complex coefficients(as multiplication is defined here)?
$(2)$ - What does restrict to $V^*$ mean? Does it mean to restrict the action of $G$ to $V^*$? I thought the action was only defined on $V^*$ to begin with?
$(3)$ - Also, when something like $mathbbC[V]$ is written in this context, what does this mean? What does $G$ acting on $mathbbC[V]$ mean?
linear-algebra abstract-algebra vector-spaces representation-theory
asked Mar 30 at 17:09
the manthe man
831716
$endgroup$
add a comment |
$begingroup$
Let $V$ be a complex vector space. Let $G$ be a finite group and let $$rho: G rightarrow GL(V)$$ be a representation of $G$.
Let $V^*$ denote the dual space of $V$, and let $mathcalO(V)$ denote the algebra of functions $F: V rightarrow mathbbC$ generated by the elements of $V^*$. Elements of $mathcalO(V)$ are called regular functions.
The dual representation $rho^*: G rightarrow GL(V^*)$ is given by $$(rho^*(g)f)(v) = f(rho(g)^-1v), forall g in G, f in V^*, v in V$$
Define the ring of invariant functions to be $$mathcalO(V)^G = f in mathcalO(V) : gf = f hspace2mm forall g in G$$
where $gf$ is just shorthand for $rho^*(g)f$.
Now, when we restrict to $V^* subset mathcalO(V)$, where $V^*$ has basis $x_1,...,x_n$, the regular functions are polynomials in the $x_i$ and the action of $G$ is given by $$gp(x_1,...,x_n) = p(gx_1,...,gx_n)$$ for some polynomial $p$.
$(1)$ - I don't understand this last bit? What is the distinction between $V^*$ and $mathcalO(V)$? Is $V^*$ just the set of $mathbbC$-linear sums of $x_1,...,x_n$? There is no multiplication between the $x_i$ defined? And an element of $mathcalO(V)$ is of the form $p(x_1,...,x_n)$ for any polynomial $p$ with complex coefficients(as multiplication is defined here)?
$(2)$ - What does restrict to $V^*$ mean? Does it mean to restrict the action of $G$ to $V^*$? I thought the action was only defined on $V^*$ to begin with?
$(3)$ - Also, when something like $mathbbC[V]$ is written in this context, what does this mean? What does $G$ acting on $mathbbC[V]$ mean?
linear-algebra abstract-algebra vector-spaces representation-theory
asked Mar 30 at 17:09
the manthe man
831716
$endgroup$
add a comment |
$begingroup$
Let $V$ be a complex vector space. Let $G$ be a finite group and let $$rho: G rightarrow GL(V)$$ be a representation of $G$.
Let $V^*$ denote the dual space of $V$, and let $mathcalO(V)$ denote the algebra of functions $F: V rightarrow mathbbC$ generated by the elements of $V^*$. Elements of $mathcalO(V)$ are called regular functions.
The dual representation $rho^*: G rightarrow GL(V^*)$ is given by $$(rho^*(g)f)(v) = f(rho(g)^-1v), forall g in G, f in V^*, v in V$$
Define the ring of invariant functions to be $$mathcalO(V)^G = f in mathcalO(V) : gf = f hspace2mm forall g in G$$
where $gf$ is just shorthand for $rho^*(g)f$.
Now, when we restrict to $V^* subset mathcalO(V)$, where $V^*$ has basis $x_1,...,x_n$, the regular functions are polynomials in the $x_i$ and the action of $G$ is given by $$gp(x_1,...,x_n) = p(gx_1,...,gx_n)$$ for some polynomial $p$.
$(1)$ - I don't understand this last bit? What is the distinction between $V^*$ and $mathcalO(V)$? Is $V^*$ just the set of $mathbbC$-linear sums of $x_1,...,x_n$? There is no multiplication between the $x_i$ defined? And an element of $mathcalO(V)$ is of the form $p(x_1,...,x_n)$ for any polynomial $p$ with complex coefficients(as multiplication is defined here)?
$(2)$ - What does restrict to $V^*$ mean? Does it mean to restrict the action of $G$ to $V^*$? I thought the action was only defined on $V^*$ to begin with?
$(3)$ - Also, when something like $mathbbC[V]$ is written in this context, what does this mean? What does $G$ acting on $mathbbC[V]$ mean?
linear-algebra abstract-algebra vector-spaces representation-theory
asked Mar 30 at 17:09
the manthe man
831716
$endgroup$
Let $V$ be a complex vector space. Let $G$ be a finite group and let $$rho: G rightarrow GL(V)$$ be a representation of $G$.
Let $V^*$ denote the dual space of $V$, and let $mathcalO(V)$ denote the algebra of functions $F: V rightarrow mathbbC$ generated by the elements of $V^*$. Elements of $mathcalO(V)$ are called regular functions.
The dual representation $rho^*: G rightarrow GL(V^*)$ is given by $$(rho^*(g)f)(v) = f(rho(g)^-1v), forall g in G, f in V^*, v in V$$
Define the ring of invariant functions to be $$mathcalO(V)^G = f in mathcalO(V) : gf = f hspace2mm forall g in G$$
where $gf$ is just shorthand for $rho^*(g)f$.
Now, when we restrict to $V^* subset mathcalO(V)$, where $V^*$ has basis $x_1,...,x_n$, the regular functions are polynomials in the $x_i$ and the action of $G$ is given by $$gp(x_1,...,x_n) = p(gx_1,...,gx_n)$$ for some polynomial $p$.
$(1)$ - I don't understand this last bit? What is the distinction between $V^*$ and $mathcalO(V)$? Is $V^*$ just the set of $mathbbC$-linear sums of $x_1,...,x_n$? There is no multiplication between the $x_i$ defined? And an element of $mathcalO(V)$ is of the form $p(x_1,...,x_n)$ for any polynomial $p$ with complex coefficients(as multiplication is defined here)?
$(2)$ - What does restrict to $V^*$ mean? Does it mean to restrict the action of $G$ to $V^*$? I thought the action was only defined on $V^*$ to begin with?
$(3)$ - Also, when something like $mathbbC[V]$ is written in this context, what does this mean? What does $G$ acting on $mathbbC[V]$ mean?
linear-algebra abstract-algebra vector-spaces representation-theory
linear-algebra abstract-algebra vector-spaces representation-theory
asked Mar 30 at 17:09
the manthe man
831716
asked Mar 30 at 17:09
the manthe man
831716
asked Mar 30 at 17:09
the manthe man
831716
asked Mar 30 at 17:09
the manthe man
831716
asked Mar 30 at 17:09
the manthe man
831716
831716
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For (1) you are completely correct.
For (2), the idea is that the statement is meant to define the action on all of $mathcalO(V)$, so the part about restriction to $V^*$ seems to be an error (since you are correct that the action was originally just defined on $V^*$).
For (3), $mathbbC[V]$ refers to the polynomial ring in $n$ variables where $n$ is the dimension of $V$, with the variables identified with a basis of $V$. But writing is like that means we don't need to pick a basis, which can have some advantages.
The main idea is that any representation on $V$ can be extended to $mathbbC[V]$ similarly to what was done for $mathcalO(V)$.
answered Mar 30 at 19:56
Tobias KildetoftTobias Kildetoft
16.9k14274
$endgroup$
$begingroup$
The action was defined on $V^*$, so how would the group act on $mathcalO(V)$? Also, how exactly is this representaion extended to $mathbbC[V]$?
$endgroup$
– the man
Mar 31 at 1:12
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
For (1) you are completely correct.
For (2), the idea is that the statement is meant to define the action on all of $mathcalO(V)$, so the part about restriction to $V^*$ seems to be an error (since you are correct that the action was originally just defined on $V^*$).
For (3), $mathbbC[V]$ refers to the polynomial ring in $n$ variables where $n$ is the dimension of $V$, with the variables identified with a basis of $V$. But writing is like that means we don't need to pick a basis, which can have some advantages.
The main idea is that any representation on $V$ can be extended to $mathbbC[V]$ similarly to what was done for $mathcalO(V)$.
answered Mar 30 at 19:56
Tobias KildetoftTobias Kildetoft
16.9k14274
$endgroup$
$begingroup$
The action was defined on $V^*$, so how would the group act on $mathcalO(V)$? Also, how exactly is this representaion extended to $mathbbC[V]$?
$endgroup$
– the man
Mar 31 at 1:12
add a comment |
$begingroup$
For (1) you are completely correct.
For (2), the idea is that the statement is meant to define the action on all of $mathcalO(V)$, so the part about restriction to $V^*$ seems to be an error (since you are correct that the action was originally just defined on $V^*$).
For (3), $mathbbC[V]$ refers to the polynomial ring in $n$ variables where $n$ is the dimension of $V$, with the variables identified with a basis of $V$. But writing is like that means we don't need to pick a basis, which can have some advantages.
The main idea is that any representation on $V$ can be extended to $mathbbC[V]$ similarly to what was done for $mathcalO(V)$.
answered Mar 30 at 19:56
Tobias KildetoftTobias Kildetoft
16.9k14274
$endgroup$
$begingroup$
The action was defined on $V^*$, so how would the group act on $mathcalO(V)$? Also, how exactly is this representaion extended to $mathbbC[V]$?
$endgroup$
– the man
Mar 31 at 1:12
add a comment |
$begingroup$
For (1) you are completely correct.
For (2), the idea is that the statement is meant to define the action on all of $mathcalO(V)$, so the part about restriction to $V^*$ seems to be an error (since you are correct that the action was originally just defined on $V^*$).
For (3), $mathbbC[V]$ refers to the polynomial ring in $n$ variables where $n$ is the dimension of $V$, with the variables identified with a basis of $V$. But writing is like that means we don't need to pick a basis, which can have some advantages.
The main idea is that any representation on $V$ can be extended to $mathbbC[V]$ similarly to what was done for $mathcalO(V)$.
answered Mar 30 at 19:56
Tobias KildetoftTobias Kildetoft
16.9k14274
$endgroup$
For (1) you are completely correct.
For (2), the idea is that the statement is meant to define the action on all of $mathcalO(V)$, so the part about restriction to $V^*$ seems to be an error (since you are correct that the action was originally just defined on $V^*$).
For (3), $mathbbC[V]$ refers to the polynomial ring in $n$ variables where $n$ is the dimension of $V$, with the variables identified with a basis of $V$. But writing is like that means we don't need to pick a basis, which can have some advantages.
The main idea is that any representation on $V$ can be extended to $mathbbC[V]$ similarly to what was done for $mathcalO(V)$.
answered Mar 30 at 19:56
Tobias KildetoftTobias Kildetoft
16.9k14274
answered Mar 30 at 19:56
Tobias KildetoftTobias Kildetoft
16.9k14274
answered Mar 30 at 19:56
Tobias KildetoftTobias Kildetoft
16.9k14274
answered Mar 30 at 19:56
Tobias KildetoftTobias Kildetoft
16.9k14274
16.9k14274
$begingroup$
The action was defined on $V^*$, so how would the group act on $mathcalO(V)$? Also, how exactly is this representaion extended to $mathbbC[V]$?
$endgroup$
– the man
Mar 31 at 1:12
add a comment |
$begingroup$
The action was defined on $V^*$, so how would the group act on $mathcalO(V)$? Also, how exactly is this representaion extended to $mathbbC[V]$?
$endgroup$
– the man
Mar 31 at 1:12
$begingroup$
The action was defined on $V^*$, so how would the group act on $mathcalO(V)$? Also, how exactly is this representaion extended to $mathbbC[V]$?
$endgroup$
– the man
Mar 31 at 1:12
$begingroup$
The action was defined on $V^*$, so how would the group act on $mathcalO(V)$? Also, how exactly is this representaion extended to $mathbbC[V]$?
$endgroup$
– the man
Mar 31 at 1:12
add a comment |
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