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Definition of symmetric power of a linear represenation
I'm reading Kowalski's Representation theory, and there's a part about the symmetric and antisymmetric powers of a representation, and I'd like to ask a question about those.
So there's a proposition in the book that says if we have a representation $rho: G to GL(E)$ and a covariant functor $T:E mapsto T(E)$, $Phi mapsto Phi_{*}$, from $Vect_{k}$ to itself, then we can construct a representation $T(rho): G to GL(T(E))$ naturally. This is obvious and I completely understand this statement.
Then, when defining the symmetric and antisymmetric powers of a representation, the author simply states that there is a covariant functor that maps $E$ to $mathrm{Sym}^{m}(E)$ (which is, as far as I understand, the set of all symmetric $m$-linear maps $f: E times dots times E to k$), and thus we can define the $m$-th symmetric power of a representation $rho: G to GL(E)$, calling it for example $mathrm{Sym}^m rho: G to GL(mathrm{Sym}^{m}(E))$. However, I don't know of any covariant functors from $E$ to $mathrm{Sym}^{m}(E)$; all I know is a contravariant functor: $$ E mapsto mathrm{Sym}^{m}(E), (Phi: E_{1} to E_{2}) mapsto (Phi^{*}: mathrm{Sym}^{m}(E_{2}) to mathrm{Sym}^{m}(E_{1}))$$
given by $Phi^{*}(f)(v_{1},...,v_{m}) = f(Phi(v_{1}),...,Phi(v_{m}))$.
My question is: What is the symmetric power of a representation?
Here are some of my observations:
- If there were a natural covariant functor from $E$ to $mathrm{Sym}^{m}(E)$, i.e. if someone knows of one, I'd appreciate a clear formulation of one, which would answer my question.
- Is there a way to define a representation with regard to a contravariant tensor? Here's my attempt in this particular case:
$$mathrm{Sym}^{m}rho: G to mathrm{Sym}^{m}(E), mathrm{Sym}^{m}rho(g)f(v_{1},...,v_{m}) = f(rho(g^{-1})v_{1},...,rho(g^{-1})v_{m}).$$ This is indeed a homomorphism, and the reason I had to put $g^{-1}$ instead of $g$ was because if I had put $g$, I'd have $mathrm{Sym}^{m}rho(gh) = mathrm{Sym}^{m}rho(h) circ mathrm{Sym}^{m}rho(g)$. Is this the standard definition?
I would appreciate any details with regards to what is standard in this case.
representation-theory functors symmetric-algebra
asked Nov 20 '18 at 14:25
Matija Sreckovic
957517
add a comment |
I'm reading Kowalski's Representation theory, and there's a part about the symmetric and antisymmetric powers of a representation, and I'd like to ask a question about those.
So there's a proposition in the book that says if we have a representation $rho: G to GL(E)$ and a covariant functor $T:E mapsto T(E)$, $Phi mapsto Phi_{*}$, from $Vect_{k}$ to itself, then we can construct a representation $T(rho): G to GL(T(E))$ naturally. This is obvious and I completely understand this statement.
Then, when defining the symmetric and antisymmetric powers of a representation, the author simply states that there is a covariant functor that maps $E$ to $mathrm{Sym}^{m}(E)$ (which is, as far as I understand, the set of all symmetric $m$-linear maps $f: E times dots times E to k$), and thus we can define the $m$-th symmetric power of a representation $rho: G to GL(E)$, calling it for example $mathrm{Sym}^m rho: G to GL(mathrm{Sym}^{m}(E))$. However, I don't know of any covariant functors from $E$ to $mathrm{Sym}^{m}(E)$; all I know is a contravariant functor: $$ E mapsto mathrm{Sym}^{m}(E), (Phi: E_{1} to E_{2}) mapsto (Phi^{*}: mathrm{Sym}^{m}(E_{2}) to mathrm{Sym}^{m}(E_{1}))$$
given by $Phi^{*}(f)(v_{1},...,v_{m}) = f(Phi(v_{1}),...,Phi(v_{m}))$.
My question is: What is the symmetric power of a representation?
Here are some of my observations:
- If there were a natural covariant functor from $E$ to $mathrm{Sym}^{m}(E)$, i.e. if someone knows of one, I'd appreciate a clear formulation of one, which would answer my question.
- Is there a way to define a representation with regard to a contravariant tensor? Here's my attempt in this particular case:
$$mathrm{Sym}^{m}rho: G to mathrm{Sym}^{m}(E), mathrm{Sym}^{m}rho(g)f(v_{1},...,v_{m}) = f(rho(g^{-1})v_{1},...,rho(g^{-1})v_{m}).$$ This is indeed a homomorphism, and the reason I had to put $g^{-1}$ instead of $g$ was because if I had put $g$, I'd have $mathrm{Sym}^{m}rho(gh) = mathrm{Sym}^{m}rho(h) circ mathrm{Sym}^{m}rho(g)$. Is this the standard definition?
I would appreciate any details with regards to what is standard in this case.
representation-theory functors symmetric-algebra
asked Nov 20 '18 at 14:25
Matija Sreckovic
957517
1
Your definition of $operatorname{Sym}^m E$ is off by a "dual", because it is actually the space where all symmetric $m$-linear maps $Etimes Etimesdotstimes Eto X$ factors through: $Etimes Etimesdotstimes Etooperatorname{Sym}^m Eto X$.
– user10354138
Nov 20 '18 at 14:41
I forgot to put "symmetric" in my definition of $mathrm{Sym}^{m}(E)$. So the set of all symmetric $m$-linear maps from $E times E times dots times E$ to $k$ is actually $(mathrm{Sym}^{m}(E))^{*}$? Could you explain to me the exact definition of $mathrm{Sym}^{m}(E)$, and how one works with that space (for example, could you name a basis)? Or even better, point me to a book which contains some clear information on the symmetric algebra?
– Matija Sreckovic
Nov 20 '18 at 15:11
2
If $x_1, ldots, x_n$ is a basis of $E$, then a standard basis of $operatorname{Sym}^m(E)$ would be ${ x_{i_1} x_{i_2} cdots x_{i_m} mid 1 le i_1 le i_2 le cdots le i_m le n }$ (where $x_{i_1} x_{i_2} cdots x_{i_m}$ denotes the image of $x_{i_1} otimes cdots otimes x_{i_m}$ in the quotient described in Qiaochu Yuan's answer.) Therefore $operatorname{Sym}^m(E)$ has dimension $binom{n+m-1}{m}$. And the symmetric algebra $S(E)$ is isomorphic to the polynomial algebra $F[x_1, ldots, x_n]$ where $F$ is the scalar field.
– Daniel Schepler
Nov 21 '18 at 1:13
add a comment |
I'm reading Kowalski's Representation theory, and there's a part about the symmetric and antisymmetric powers of a representation, and I'd like to ask a question about those.
So there's a proposition in the book that says if we have a representation $rho: G to GL(E)$ and a covariant functor $T:E mapsto T(E)$, $Phi mapsto Phi_{*}$, from $Vect_{k}$ to itself, then we can construct a representation $T(rho): G to GL(T(E))$ naturally. This is obvious and I completely understand this statement.
Then, when defining the symmetric and antisymmetric powers of a representation, the author simply states that there is a covariant functor that maps $E$ to $mathrm{Sym}^{m}(E)$ (which is, as far as I understand, the set of all symmetric $m$-linear maps $f: E times dots times E to k$), and thus we can define the $m$-th symmetric power of a representation $rho: G to GL(E)$, calling it for example $mathrm{Sym}^m rho: G to GL(mathrm{Sym}^{m}(E))$. However, I don't know of any covariant functors from $E$ to $mathrm{Sym}^{m}(E)$; all I know is a contravariant functor: $$ E mapsto mathrm{Sym}^{m}(E), (Phi: E_{1} to E_{2}) mapsto (Phi^{*}: mathrm{Sym}^{m}(E_{2}) to mathrm{Sym}^{m}(E_{1}))$$
given by $Phi^{*}(f)(v_{1},...,v_{m}) = f(Phi(v_{1}),...,Phi(v_{m}))$.
My question is: What is the symmetric power of a representation?
Here are some of my observations:
- If there were a natural covariant functor from $E$ to $mathrm{Sym}^{m}(E)$, i.e. if someone knows of one, I'd appreciate a clear formulation of one, which would answer my question.
- Is there a way to define a representation with regard to a contravariant tensor? Here's my attempt in this particular case:
$$mathrm{Sym}^{m}rho: G to mathrm{Sym}^{m}(E), mathrm{Sym}^{m}rho(g)f(v_{1},...,v_{m}) = f(rho(g^{-1})v_{1},...,rho(g^{-1})v_{m}).$$ This is indeed a homomorphism, and the reason I had to put $g^{-1}$ instead of $g$ was because if I had put $g$, I'd have $mathrm{Sym}^{m}rho(gh) = mathrm{Sym}^{m}rho(h) circ mathrm{Sym}^{m}rho(g)$. Is this the standard definition?
I would appreciate any details with regards to what is standard in this case.
representation-theory functors symmetric-algebra
asked Nov 20 '18 at 14:25
Matija Sreckovic
957517
I'm reading Kowalski's Representation theory, and there's a part about the symmetric and antisymmetric powers of a representation, and I'd like to ask a question about those.
So there's a proposition in the book that says if we have a representation $rho: G to GL(E)$ and a covariant functor $T:E mapsto T(E)$, $Phi mapsto Phi_{*}$, from $Vect_{k}$ to itself, then we can construct a representation $T(rho): G to GL(T(E))$ naturally. This is obvious and I completely understand this statement.
Then, when defining the symmetric and antisymmetric powers of a representation, the author simply states that there is a covariant functor that maps $E$ to $mathrm{Sym}^{m}(E)$ (which is, as far as I understand, the set of all symmetric $m$-linear maps $f: E times dots times E to k$), and thus we can define the $m$-th symmetric power of a representation $rho: G to GL(E)$, calling it for example $mathrm{Sym}^m rho: G to GL(mathrm{Sym}^{m}(E))$. However, I don't know of any covariant functors from $E$ to $mathrm{Sym}^{m}(E)$; all I know is a contravariant functor: $$ E mapsto mathrm{Sym}^{m}(E), (Phi: E_{1} to E_{2}) mapsto (Phi^{*}: mathrm{Sym}^{m}(E_{2}) to mathrm{Sym}^{m}(E_{1}))$$
given by $Phi^{*}(f)(v_{1},...,v_{m}) = f(Phi(v_{1}),...,Phi(v_{m}))$.
My question is: What is the symmetric power of a representation?
Here are some of my observations:
- If there were a natural covariant functor from $E$ to $mathrm{Sym}^{m}(E)$, i.e. if someone knows of one, I'd appreciate a clear formulation of one, which would answer my question.
- Is there a way to define a representation with regard to a contravariant tensor? Here's my attempt in this particular case:
$$mathrm{Sym}^{m}rho: G to mathrm{Sym}^{m}(E), mathrm{Sym}^{m}rho(g)f(v_{1},...,v_{m}) = f(rho(g^{-1})v_{1},...,rho(g^{-1})v_{m}).$$ This is indeed a homomorphism, and the reason I had to put $g^{-1}$ instead of $g$ was because if I had put $g$, I'd have $mathrm{Sym}^{m}rho(gh) = mathrm{Sym}^{m}rho(h) circ mathrm{Sym}^{m}rho(g)$. Is this the standard definition?
I would appreciate any details with regards to what is standard in this case.
representation-theory functors symmetric-algebra
representation-theory functors symmetric-algebra
asked Nov 20 '18 at 14:25
Matija Sreckovic
957517
asked Nov 20 '18 at 14:25
Matija Sreckovic
957517
edited Nov 20 '18 at 15:21
asked Nov 20 '18 at 14:25
Matija Sreckovic
957517
asked Nov 20 '18 at 14:25
Matija Sreckovic
957517
asked Nov 20 '18 at 14:25
Matija Sreckovic
957517
957517
1
Your definition of $operatorname{Sym}^m E$ is off by a "dual", because it is actually the space where all symmetric $m$-linear maps $Etimes Etimesdotstimes Eto X$ factors through: $Etimes Etimesdotstimes Etooperatorname{Sym}^m Eto X$.
– user10354138
Nov 20 '18 at 14:41
I forgot to put "symmetric" in my definition of $mathrm{Sym}^{m}(E)$. So the set of all symmetric $m$-linear maps from $E times E times dots times E$ to $k$ is actually $(mathrm{Sym}^{m}(E))^{*}$? Could you explain to me the exact definition of $mathrm{Sym}^{m}(E)$, and how one works with that space (for example, could you name a basis)? Or even better, point me to a book which contains some clear information on the symmetric algebra?
– Matija Sreckovic
Nov 20 '18 at 15:11
2
If $x_1, ldots, x_n$ is a basis of $E$, then a standard basis of $operatorname{Sym}^m(E)$ would be ${ x_{i_1} x_{i_2} cdots x_{i_m} mid 1 le i_1 le i_2 le cdots le i_m le n }$ (where $x_{i_1} x_{i_2} cdots x_{i_m}$ denotes the image of $x_{i_1} otimes cdots otimes x_{i_m}$ in the quotient described in Qiaochu Yuan's answer.) Therefore $operatorname{Sym}^m(E)$ has dimension $binom{n+m-1}{m}$. And the symmetric algebra $S(E)$ is isomorphic to the polynomial algebra $F[x_1, ldots, x_n]$ where $F$ is the scalar field.
– Daniel Schepler
Nov 21 '18 at 1:13
add a comment |
1
Your definition of $operatorname{Sym}^m E$ is off by a "dual", because it is actually the space where all symmetric $m$-linear maps $Etimes Etimesdotstimes Eto X$ factors through: $Etimes Etimesdotstimes Etooperatorname{Sym}^m Eto X$.
– user10354138
Nov 20 '18 at 14:41
I forgot to put "symmetric" in my definition of $mathrm{Sym}^{m}(E)$. So the set of all symmetric $m$-linear maps from $E times E times dots times E$ to $k$ is actually $(mathrm{Sym}^{m}(E))^{*}$? Could you explain to me the exact definition of $mathrm{Sym}^{m}(E)$, and how one works with that space (for example, could you name a basis)? Or even better, point me to a book which contains some clear information on the symmetric algebra?
– Matija Sreckovic
Nov 20 '18 at 15:11
2
If $x_1, ldots, x_n$ is a basis of $E$, then a standard basis of $operatorname{Sym}^m(E)$ would be ${ x_{i_1} x_{i_2} cdots x_{i_m} mid 1 le i_1 le i_2 le cdots le i_m le n }$ (where $x_{i_1} x_{i_2} cdots x_{i_m}$ denotes the image of $x_{i_1} otimes cdots otimes x_{i_m}$ in the quotient described in Qiaochu Yuan's answer.) Therefore $operatorname{Sym}^m(E)$ has dimension $binom{n+m-1}{m}$. And the symmetric algebra $S(E)$ is isomorphic to the polynomial algebra $F[x_1, ldots, x_n]$ where $F$ is the scalar field.
– Daniel Schepler
Nov 21 '18 at 1:13
1
1
Your definition of $operatorname{Sym}^m E$ is off by a "dual", because it is actually the space where all symmetric $m$-linear maps $Etimes Etimesdotstimes Eto X$ factors through: $Etimes Etimesdotstimes Etooperatorname{Sym}^m Eto X$.
– user10354138
Nov 20 '18 at 14:41
Your definition of $operatorname{Sym}^m E$ is off by a "dual", because it is actually the space where all symmetric $m$-linear maps $Etimes Etimesdotstimes Eto X$ factors through: $Etimes Etimesdotstimes Etooperatorname{Sym}^m Eto X$.
– user10354138
Nov 20 '18 at 14:41
I forgot to put "symmetric" in my definition of $mathrm{Sym}^{m}(E)$. So the set of all symmetric $m$-linear maps from $E times E times dots times E$ to $k$ is actually $(mathrm{Sym}^{m}(E))^{*}$? Could you explain to me the exact definition of $mathrm{Sym}^{m}(E)$, and how one works with that space (for example, could you name a basis)? Or even better, point me to a book which contains some clear information on the symmetric algebra?
– Matija Sreckovic
Nov 20 '18 at 15:11
I forgot to put "symmetric" in my definition of $mathrm{Sym}^{m}(E)$. So the set of all symmetric $m$-linear maps from $E times E times dots times E$ to $k$ is actually $(mathrm{Sym}^{m}(E))^{*}$? Could you explain to me the exact definition of $mathrm{Sym}^{m}(E)$, and how one works with that space (for example, could you name a basis)? Or even better, point me to a book which contains some clear information on the symmetric algebra?
– Matija Sreckovic
Nov 20 '18 at 15:11
2
2
If $x_1, ldots, x_n$ is a basis of $E$, then a standard basis of $operatorname{Sym}^m(E)$ would be ${ x_{i_1} x_{i_2} cdots x_{i_m} mid 1 le i_1 le i_2 le cdots le i_m le n }$ (where $x_{i_1} x_{i_2} cdots x_{i_m}$ denotes the image of $x_{i_1} otimes cdots otimes x_{i_m}$ in the quotient described in Qiaochu Yuan's answer.) Therefore $operatorname{Sym}^m(E)$ has dimension $binom{n+m-1}{m}$. And the symmetric algebra $S(E)$ is isomorphic to the polynomial algebra $F[x_1, ldots, x_n]$ where $F$ is the scalar field.
– Daniel Schepler
Nov 21 '18 at 1:13
If $x_1, ldots, x_n$ is a basis of $E$, then a standard basis of $operatorname{Sym}^m(E)$ would be ${ x_{i_1} x_{i_2} cdots x_{i_m} mid 1 le i_1 le i_2 le cdots le i_m le n }$ (where $x_{i_1} x_{i_2} cdots x_{i_m}$ denotes the image of $x_{i_1} otimes cdots otimes x_{i_m}$ in the quotient described in Qiaochu Yuan's answer.) Therefore $operatorname{Sym}^m(E)$ has dimension $binom{n+m-1}{m}$. And the symmetric algebra $S(E)$ is isomorphic to the polynomial algebra $F[x_1, ldots, x_n]$ where $F$ is the scalar field.
– Daniel Schepler
Nov 21 '18 at 1:13
add a comment |
1 Answer
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What you've defined is not the symmetric power, since as you say it's contravariant. The $n^{th}$ symmetric power $S^n(V)$ of a vector space $V$ is the quotient of the $n^{th}$ tensor power $V^{otimes n}$ (which is also covariant) by the action of the symmetric group $S_n$ permuting the factors. It has the following universal property: every symmetric multilinear map $V^n to W$ factors uniquely through a linear map $S^n(V) to W$. So there is a universal symmetric multilinear map
$$V^n to S^n(V).$$
What you call the symmetric power is the dual of $S^n(V)$.
The symmetric powers are the graded components of the symmetric algebra
$$S(V) = bigoplus_{n ge 0} S^n(V)$$
and in particular there's a bilinear map $S^n(V) times S^m(V) to S^{n+m}(V)$ giving the symmetric algebra the structure of a commutative algebra. If $e_1, dots e_n$ is a basis for $V$, $S^n(V)$ has a basis consisting of all monomials in the $e_i$, thought of as variables, of degree $n$, and $S(V)$ is the polynomial algebra on the $e_i$.
As for your second question, yes, and that's how you do it. See dual representation.
answered Nov 21 '18 at 0:44
Qiaochu Yuan
277k32581919
add a comment |
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What you've defined is not the symmetric power, since as you say it's contravariant. The $n^{th}$ symmetric power $S^n(V)$ of a vector space $V$ is the quotient of the $n^{th}$ tensor power $V^{otimes n}$ (which is also covariant) by the action of the symmetric group $S_n$ permuting the factors. It has the following universal property: every symmetric multilinear map $V^n to W$ factors uniquely through a linear map $S^n(V) to W$. So there is a universal symmetric multilinear map
$$V^n to S^n(V).$$
What you call the symmetric power is the dual of $S^n(V)$.
The symmetric powers are the graded components of the symmetric algebra
$$S(V) = bigoplus_{n ge 0} S^n(V)$$
and in particular there's a bilinear map $S^n(V) times S^m(V) to S^{n+m}(V)$ giving the symmetric algebra the structure of a commutative algebra. If $e_1, dots e_n$ is a basis for $V$, $S^n(V)$ has a basis consisting of all monomials in the $e_i$, thought of as variables, of degree $n$, and $S(V)$ is the polynomial algebra on the $e_i$.
As for your second question, yes, and that's how you do it. See dual representation.
answered Nov 21 '18 at 0:44
Qiaochu Yuan
277k32581919
add a comment |
What you've defined is not the symmetric power, since as you say it's contravariant. The $n^{th}$ symmetric power $S^n(V)$ of a vector space $V$ is the quotient of the $n^{th}$ tensor power $V^{otimes n}$ (which is also covariant) by the action of the symmetric group $S_n$ permuting the factors. It has the following universal property: every symmetric multilinear map $V^n to W$ factors uniquely through a linear map $S^n(V) to W$. So there is a universal symmetric multilinear map
$$V^n to S^n(V).$$
What you call the symmetric power is the dual of $S^n(V)$.
The symmetric powers are the graded components of the symmetric algebra
$$S(V) = bigoplus_{n ge 0} S^n(V)$$
and in particular there's a bilinear map $S^n(V) times S^m(V) to S^{n+m}(V)$ giving the symmetric algebra the structure of a commutative algebra. If $e_1, dots e_n$ is a basis for $V$, $S^n(V)$ has a basis consisting of all monomials in the $e_i$, thought of as variables, of degree $n$, and $S(V)$ is the polynomial algebra on the $e_i$.
As for your second question, yes, and that's how you do it. See dual representation.
answered Nov 21 '18 at 0:44
Qiaochu Yuan
277k32581919
add a comment |
What you've defined is not the symmetric power, since as you say it's contravariant. The $n^{th}$ symmetric power $S^n(V)$ of a vector space $V$ is the quotient of the $n^{th}$ tensor power $V^{otimes n}$ (which is also covariant) by the action of the symmetric group $S_n$ permuting the factors. It has the following universal property: every symmetric multilinear map $V^n to W$ factors uniquely through a linear map $S^n(V) to W$. So there is a universal symmetric multilinear map
$$V^n to S^n(V).$$
What you call the symmetric power is the dual of $S^n(V)$.
The symmetric powers are the graded components of the symmetric algebra
$$S(V) = bigoplus_{n ge 0} S^n(V)$$
and in particular there's a bilinear map $S^n(V) times S^m(V) to S^{n+m}(V)$ giving the symmetric algebra the structure of a commutative algebra. If $e_1, dots e_n$ is a basis for $V$, $S^n(V)$ has a basis consisting of all monomials in the $e_i$, thought of as variables, of degree $n$, and $S(V)$ is the polynomial algebra on the $e_i$.
As for your second question, yes, and that's how you do it. See dual representation.
answered Nov 21 '18 at 0:44
Qiaochu Yuan
277k32581919
What you've defined is not the symmetric power, since as you say it's contravariant. The $n^{th}$ symmetric power $S^n(V)$ of a vector space $V$ is the quotient of the $n^{th}$ tensor power $V^{otimes n}$ (which is also covariant) by the action of the symmetric group $S_n$ permuting the factors. It has the following universal property: every symmetric multilinear map $V^n to W$ factors uniquely through a linear map $S^n(V) to W$. So there is a universal symmetric multilinear map
$$V^n to S^n(V).$$
What you call the symmetric power is the dual of $S^n(V)$.
The symmetric powers are the graded components of the symmetric algebra
$$S(V) = bigoplus_{n ge 0} S^n(V)$$
and in particular there's a bilinear map $S^n(V) times S^m(V) to S^{n+m}(V)$ giving the symmetric algebra the structure of a commutative algebra. If $e_1, dots e_n$ is a basis for $V$, $S^n(V)$ has a basis consisting of all monomials in the $e_i$, thought of as variables, of degree $n$, and $S(V)$ is the polynomial algebra on the $e_i$.
As for your second question, yes, and that's how you do it. See dual representation.
answered Nov 21 '18 at 0:44
Qiaochu Yuan
277k32581919
answered Nov 21 '18 at 0:44
Qiaochu Yuan
277k32581919
answered Nov 21 '18 at 0:44
Qiaochu Yuan
277k32581919
answered Nov 21 '18 at 0:44
Qiaochu Yuan
277k32581919
277k32581919
add a comment |
add a comment |
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1
Your definition of $operatorname{Sym}^m E$ is off by a "dual", because it is actually the space where all symmetric $m$-linear maps $Etimes Etimesdotstimes Eto X$ factors through: $Etimes Etimesdotstimes Etooperatorname{Sym}^m Eto X$.
– user10354138
Nov 20 '18 at 14:41
I forgot to put "symmetric" in my definition of $mathrm{Sym}^{m}(E)$. So the set of all symmetric $m$-linear maps from $E times E times dots times E$ to $k$ is actually $(mathrm{Sym}^{m}(E))^{*}$? Could you explain to me the exact definition of $mathrm{Sym}^{m}(E)$, and how one works with that space (for example, could you name a basis)? Or even better, point me to a book which contains some clear information on the symmetric algebra?
– Matija Sreckovic
Nov 20 '18 at 15:11
2
If $x_1, ldots, x_n$ is a basis of $E$, then a standard basis of $operatorname{Sym}^m(E)$ would be ${ x_{i_1} x_{i_2} cdots x_{i_m} mid 1 le i_1 le i_2 le cdots le i_m le n }$ (where $x_{i_1} x_{i_2} cdots x_{i_m}$ denotes the image of $x_{i_1} otimes cdots otimes x_{i_m}$ in the quotient described in Qiaochu Yuan's answer.) Therefore $operatorname{Sym}^m(E)$ has dimension $binom{n+m-1}{m}$. And the symmetric algebra $S(E)$ is isomorphic to the polynomial algebra $F[x_1, ldots, x_n]$ where $F$ is the scalar field.
– Daniel Schepler
Nov 21 '18 at 1:13