Mixing
Symmetric Algebra over Tensor Product
Let $k$ be a field. I am interested in the symmetric algebra functor $S : k text{-vect} rightarrow k text{-alg}$ taking a $k$-vector space $V$ to the symmetric algebra $S(V)$ over $V$, which is a quotient of the tensor algebra $T(V) / I$, where $I$ consists of the ideal generated by tensors of the form $x otimes y - y otimes x$ for $x, y in V$. $S$ is a left adjoint functor.
$S(V)$ is sort of like an exponential map, in my intuition. I have far reaching but informal reasons for saying this. For the space I'm given it's hard to develop the analogy, though maybe someone here can make this formal.
1) $S$ sends $0$ to $k$. (like how the exponential map sends $0$ to $1$.)
2) $S$ sends $k$ to $k(x)$. (like how the exponential map sends $0$ to $e$).
3) $S$ sends $k^n$ to $k(x_1) otimes_k cdots otimes_k k(x_n)$. (like how the exponential map sends $n$ to $e^n$).
4) $S$ sends $V oplus W$ to $S(V) otimes_k S(W)$ (like how the exponential map sends $a+b$ to $e^a cdot e^b$).
Now my question:
5) $S$ sends $V otimes_k W$ to what? In other words, what is a natural choice of a functor $k text{-alg} times k text{-alg} rightarrow k text{-alg}$ sends $(S(V) ,S(W))$ to $S(V otimes W)$. This would sort of be analogous to $(e^a, e^b) mapsto e^{ab}$. We know to define it on free algebras. Maybe that means we can define it on quotients of free algebras using this formula.
tensor-products algebras
asked Nov 21 '18 at 22:04
Dean YoungDean Young
1,474720
add a comment |
Let $k$ be a field. I am interested in the symmetric algebra functor $S : k text{-vect} rightarrow k text{-alg}$ taking a $k$-vector space $V$ to the symmetric algebra $S(V)$ over $V$, which is a quotient of the tensor algebra $T(V) / I$, where $I$ consists of the ideal generated by tensors of the form $x otimes y - y otimes x$ for $x, y in V$. $S$ is a left adjoint functor.
$S(V)$ is sort of like an exponential map, in my intuition. I have far reaching but informal reasons for saying this. For the space I'm given it's hard to develop the analogy, though maybe someone here can make this formal.
1) $S$ sends $0$ to $k$. (like how the exponential map sends $0$ to $1$.)
2) $S$ sends $k$ to $k(x)$. (like how the exponential map sends $0$ to $e$).
3) $S$ sends $k^n$ to $k(x_1) otimes_k cdots otimes_k k(x_n)$. (like how the exponential map sends $n$ to $e^n$).
4) $S$ sends $V oplus W$ to $S(V) otimes_k S(W)$ (like how the exponential map sends $a+b$ to $e^a cdot e^b$).
Now my question:
5) $S$ sends $V otimes_k W$ to what? In other words, what is a natural choice of a functor $k text{-alg} times k text{-alg} rightarrow k text{-alg}$ sends $(S(V) ,S(W))$ to $S(V otimes W)$. This would sort of be analogous to $(e^a, e^b) mapsto e^{ab}$. We know to define it on free algebras. Maybe that means we can define it on quotients of free algebras using this formula.
tensor-products algebras
asked Nov 21 '18 at 22:04
Dean YoungDean Young
1,474720
If $mathrm{dim}(V) = n$ and $mathrm{dim}(W) = m$, then $S(V otimes W)$ is a polynomial ring in $nm$ variables, because $mathrm{dim}(V otimes W) = nm$.
– Nick
Nov 21 '18 at 23:08
Yes, but of course this is not the complete picture since not all $k$-algebras are free. i.e. we want a natural choice of functor whose domain is $k text{-alg} times k text{-alg}$ and whose codomain is $k text{-alg}$.
– Dean Young
Nov 21 '18 at 23:19
add a comment |
Let $k$ be a field. I am interested in the symmetric algebra functor $S : k text{-vect} rightarrow k text{-alg}$ taking a $k$-vector space $V$ to the symmetric algebra $S(V)$ over $V$, which is a quotient of the tensor algebra $T(V) / I$, where $I$ consists of the ideal generated by tensors of the form $x otimes y - y otimes x$ for $x, y in V$. $S$ is a left adjoint functor.
$S(V)$ is sort of like an exponential map, in my intuition. I have far reaching but informal reasons for saying this. For the space I'm given it's hard to develop the analogy, though maybe someone here can make this formal.
1) $S$ sends $0$ to $k$. (like how the exponential map sends $0$ to $1$.)
2) $S$ sends $k$ to $k(x)$. (like how the exponential map sends $0$ to $e$).
3) $S$ sends $k^n$ to $k(x_1) otimes_k cdots otimes_k k(x_n)$. (like how the exponential map sends $n$ to $e^n$).
4) $S$ sends $V oplus W$ to $S(V) otimes_k S(W)$ (like how the exponential map sends $a+b$ to $e^a cdot e^b$).
Now my question:
5) $S$ sends $V otimes_k W$ to what? In other words, what is a natural choice of a functor $k text{-alg} times k text{-alg} rightarrow k text{-alg}$ sends $(S(V) ,S(W))$ to $S(V otimes W)$. This would sort of be analogous to $(e^a, e^b) mapsto e^{ab}$. We know to define it on free algebras. Maybe that means we can define it on quotients of free algebras using this formula.
tensor-products algebras
asked Nov 21 '18 at 22:04
Dean YoungDean Young
1,474720
Let $k$ be a field. I am interested in the symmetric algebra functor $S : k text{-vect} rightarrow k text{-alg}$ taking a $k$-vector space $V$ to the symmetric algebra $S(V)$ over $V$, which is a quotient of the tensor algebra $T(V) / I$, where $I$ consists of the ideal generated by tensors of the form $x otimes y - y otimes x$ for $x, y in V$. $S$ is a left adjoint functor.
$S(V)$ is sort of like an exponential map, in my intuition. I have far reaching but informal reasons for saying this. For the space I'm given it's hard to develop the analogy, though maybe someone here can make this formal.
1) $S$ sends $0$ to $k$. (like how the exponential map sends $0$ to $1$.)
2) $S$ sends $k$ to $k(x)$. (like how the exponential map sends $0$ to $e$).
3) $S$ sends $k^n$ to $k(x_1) otimes_k cdots otimes_k k(x_n)$. (like how the exponential map sends $n$ to $e^n$).
4) $S$ sends $V oplus W$ to $S(V) otimes_k S(W)$ (like how the exponential map sends $a+b$ to $e^a cdot e^b$).
Now my question:
5) $S$ sends $V otimes_k W$ to what? In other words, what is a natural choice of a functor $k text{-alg} times k text{-alg} rightarrow k text{-alg}$ sends $(S(V) ,S(W))$ to $S(V otimes W)$. This would sort of be analogous to $(e^a, e^b) mapsto e^{ab}$. We know to define it on free algebras. Maybe that means we can define it on quotients of free algebras using this formula.
tensor-products algebras
tensor-products algebras
asked Nov 21 '18 at 22:04
Dean YoungDean Young
1,474720
asked Nov 21 '18 at 22:04
Dean YoungDean Young
1,474720
asked Nov 21 '18 at 22:04
Dean YoungDean Young
1,474720
asked Nov 21 '18 at 22:04
Dean YoungDean Young
1,474720
asked Nov 21 '18 at 22:04
Dean YoungDean Young
1,474720
1,474720
If $mathrm{dim}(V) = n$ and $mathrm{dim}(W) = m$, then $S(V otimes W)$ is a polynomial ring in $nm$ variables, because $mathrm{dim}(V otimes W) = nm$.
– Nick
Nov 21 '18 at 23:08
Yes, but of course this is not the complete picture since not all $k$-algebras are free. i.e. we want a natural choice of functor whose domain is $k text{-alg} times k text{-alg}$ and whose codomain is $k text{-alg}$.
– Dean Young
Nov 21 '18 at 23:19
add a comment |
If $mathrm{dim}(V) = n$ and $mathrm{dim}(W) = m$, then $S(V otimes W)$ is a polynomial ring in $nm$ variables, because $mathrm{dim}(V otimes W) = nm$.
– Nick
Nov 21 '18 at 23:08
Yes, but of course this is not the complete picture since not all $k$-algebras are free. i.e. we want a natural choice of functor whose domain is $k text{-alg} times k text{-alg}$ and whose codomain is $k text{-alg}$.
– Dean Young
Nov 21 '18 at 23:19
If $mathrm{dim}(V) = n$ and $mathrm{dim}(W) = m$, then $S(V otimes W)$ is a polynomial ring in $nm$ variables, because $mathrm{dim}(V otimes W) = nm$.
– Nick
Nov 21 '18 at 23:08
If $mathrm{dim}(V) = n$ and $mathrm{dim}(W) = m$, then $S(V otimes W)$ is a polynomial ring in $nm$ variables, because $mathrm{dim}(V otimes W) = nm$.
– Nick
Nov 21 '18 at 23:08
Yes, but of course this is not the complete picture since not all $k$-algebras are free. i.e. we want a natural choice of functor whose domain is $k text{-alg} times k text{-alg}$ and whose codomain is $k text{-alg}$.
– Dean Young
Nov 21 '18 at 23:19
Yes, but of course this is not the complete picture since not all $k$-algebras are free. i.e. we want a natural choice of functor whose domain is $k text{-alg} times k text{-alg}$ and whose codomain is $k text{-alg}$.
– Dean Young
Nov 21 '18 at 23:19
add a comment |
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If $mathrm{dim}(V) = n$ and $mathrm{dim}(W) = m$, then $S(V otimes W)$ is a polynomial ring in $nm$ variables, because $mathrm{dim}(V otimes W) = nm$.
– Nick
Nov 21 '18 at 23:08
Yes, but of course this is not the complete picture since not all $k$-algebras are free. i.e. we want a natural choice of functor whose domain is $k text{-alg} times k text{-alg}$ and whose codomain is $k text{-alg}$.
– Dean Young
Nov 21 '18 at 23:19