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Let's work over an algebraically closed field and let $V$ be a finite dimensional vector space. Then is there a natural chain of maps $$ (V^*)^{\otimes l} \stackrel{f}{\rightarrow} S^l( (V^*)^{\oplus l} )\cong S^l( (V^{\oplus l})^* ) \stackrel{g}{\rightarrow} S^l(V^*), $$ where $S^l(W)$ is the symmetric power of a finite dimensional vector space $W$ and $W^*$ means the set of linear functionals on $W$?

Note that $g$ is supposed to be induced by the natural diagonal embedding $V\hookrightarrow V^{\oplus l}$ and note $S^l( (V^*)^{\oplus l})=\oplus_{p_1+\ldots +p_l=l} \otimes_{i=1}^l S^{p_i}(V^*)$.

When $l=2$, the above simplifies as: $$ (V^*)^{\otimes 2}\stackrel{f}{\rightarrow} S^0(V^*)\otimes S^2(V^*)\oplus S^1(V^*)\otimes S^1(V^*)\oplus S^2(V^*)\otimes S^0(V^*)\stackrel{g}{\rightarrow}S^2(V^*). $$ Then does $f$ send $f_1\otimes f_2\mapsto f_1 f_2 \oplus (\sum_{\sigma\in S_2} f_{\sigma(1)}\otimes f_{\sigma(2)}) \oplus f_1 f_2$, and then does $g$ send $f_1 f_2 \oplus f_1\otimes f_2 \oplus f_1 f_2\mapsto f_1f_2$? If the sequence of maps that I wrote down is correct, then what is the point of embedding $(V^*)^{\otimes l}$ through $f$ into the symmetric power of a larger vector space?

user85336
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Let us discuss the $l=2$ case. The point is that

$$V=S^1(V):=T(V)/(I_V),$$

where $I_V$ is the two sided ideal generated by $v_1\otimes v_2-v_2\otimes v_1$, for all $v_1, v_2\in V$. Analogously, we have the embedding (or the definition)

$$V^{*}\otimes V^{*}=S^1(V^{*}\otimes V^{*}).$$

If we want to embed $V^{*}\otimes V^{*}$ into the weight $2$ component $S^2(\cdot)$ of a symmetric algebra, we need to "double the dimension". Let us do it. First of all, in what follows we use the injections $W\hookrightarrow W\oplus W$ on the first resp. second factor given by $w\mapsto w\oplus 1$, resp. $w\mapsto 1\oplus w$. The morphism

$$V^{*}\otimes V^{*}\stackrel{f}{\rightarrow} S^2(V^{*}\oplus V^{*})$$

is given by the composition

$$V^{*}\otimes V^{*}\rightarrow (V^{*}\oplus V^{*})\otimes (V^{*}\oplus V^{*})\twoheadrightarrow S^2(V^{*}\oplus V^{*})=((V^{*}\oplus V^{*})\otimes (V^{*}\oplus V^{*}))/I_{V^{*}\oplus V^{*}} $$

with

$$\epsilon_1\otimes \epsilon_2 \mapsto (\epsilon_1\oplus 1)\otimes (1\oplus \epsilon_2) + I_{V^{*}\oplus V^{*}},$$

for any basis $\{\epsilon_i\}$ of $V^{*}$. The isomorphism

$$S^2(V^{*}\oplus V^{*})\stackrel{\mathcal I}{\rightarrow} S^2((V\oplus V)^{*})= ((V\oplus V)^{*}\otimes (V\oplus V)^{*})/I_{(V\oplus V)^{*}}$$

is given by (on the image of $f$)

$$(\epsilon_1\oplus 1)\otimes (1\oplus \epsilon_2) + I_{V^{*}\oplus V^{*}}\mapsto (e_1\oplus 1)\otimes (1\oplus e_2) + I_{(V\oplus V)^{*}}, $$

denoting by $\{e_i\}$ the dual basis in $V$, i.e. $\epsilon_i(e_j)=\delta_{ij}$.

We arrive at the morphism $g$

$$S^2((V\oplus V)^{*})\stackrel{g}{\rightarrow} S^2(V^{*})= (V^{*}\otimes V^{*})/I_{V^{*}}$$

given by (on the image of $\mathcal I\circ f$)

$$(e_1\oplus 1)\otimes (1\oplus e_2) + I_{(V\oplus V)^{*}}\mapsto (\epsilon_1\otimes \epsilon_2) + I_{V^{*}},$$

and we are done.

Avitus
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