What are the algebraic features and the geometric interpretation of the symmetric algebra?

Question

The exterior algebra is a quotient of the tensor algebra that gives an anti symmetric product. The symmetric algebra is similar except that the product is symmetric. The objects in both algebras are not tensors. I know that the exterior algebra consists of oriented plane segments, but what are the algebraic features and the geometric interpretation of the symmetric algebra?

Answer 1

Question: "I know that the exterior algebra consists of oriented plane segments, but what are the algebraic features and the geometric interpretation of the symmetric algebra?"

Answer: The symmetric algebra $Sym_A^*(E^*)$ of the dual of a left $A$-module $E$ ($A$ is a commutative unital ring) is the unique $A$-algebra satisfying the following universal property: There is for every commutative unital $A$-algebra $R$ a canonical isomorphism

$$Hom_{A-mod}(E^*,R) \cong Hom_{A-alg}(Sym_A^*(E^*),R)$$

In algebra/algebraic geometry one use this construction to construct various algebraic vector bundles such as the cotangent bundle, the tangent bundle. If $\Omega^1_{A/k}$ is the module of Kahler differentials of $A/k$, by definition

$$T^*(S):=\mathbb{V}((\Omega^1_{A/k})^*):=Spec(Sym_A^*((\Omega^1_{A/k})^*))$$

is the cotangent bundle of $S:=Spec(A)$.

Note: If $\Omega^1_{A/k}$ is a finite rank locally trivial $A$-module, it follows the canonical map $\pi: \mathbb{V}((\Omega^1_{A/k})^*)\rightarrow S$ is locally trivial in the Zariski topology: There is an open cover $\cup_i U_i:=\cup_i Spec(B_i) = S$ with $\pi^{-1}(U_i) \cong U_i \times_k \mathbb{A}^d_k$. A local trivialization $B_i \otimes_A E \cong B_i\{e_1,..,e_d\}$ gives an isomorphism

$$B_i\otimes_A Sym_A^*(E^*) \cong Sym_{B_i}^*((B_i\otimes_A E)^*) \cong B_i[x_1,..,x_d]$$

with $x_i:=e_i^*$ the dual basis. Hence

$$Spec(B_i\otimes_A Sym_A^*(E^*)) \cong Spec(B_i[x_1,..,x_n])\cong U_i\times_k \mathbb{A}^d_k.$$

There is an exercise in Hartshorne (HH.Ex.II.5.18) proving that the above construction gives an "equivalence of categories" between the category of finite rank geometric vector bundles on $Spec(A)$ and the category of finite rank projective $A$-modules. This is similar to the "Serre-Swan theorem" in differential geometry giving a relation between a smooth vector bundle and its module of smooth sections. Hence when people are studying $A$-modules and projective $A$-modules, they are studying "geometric vector bundles" which are geometric objects.

Note 1: There are three notions that are related: An $A$-module $E$ can be $1.$ A rank $n$ projective $A$-module. $2.$ Locally trivial in the sense that there is an open cover $D(f_i)$ of $S:=Spec(A)$ with $E_{f_i} \cong A_{f_i}^n$ for all $i$. $3.$ For any prime ideal $\mathfrak{p} \subseteq A$ it follows $E_{\mathfrak{p}} \cong A_{\mathfrak{p}}^n$. These 3 notions are related but not equivalent in general.

Note 2: If $\mathcal{E}:=\tilde{E}$ is the $\mathcal{O}_S$-module of $E$ it follows the stalk of $\mathcal{E}$ at $\mathfrak{p}$ is

$$\mathcal{E}_{\mathfrak{p}} \cong E_{\mathfrak{p}}.$$

There is an isomorphism

$$\mathcal{E}(D(f_i)) \cong E_{f_i} \cong A_{f_i}^n \cong \mathcal{O}_S(D(f_i))^n$$

hence if $E$ is locally trivial of rank $n$ in the above sense, it follows the sheaf $\mathcal{E}$ is a locally trivial $\mathcal{O}_S$-module of rank $n$.

If $E$ is projective it follows the sections of the map $\pi: Spec(Sym_A^*(E^*)) \rightarrow S$ are in 1-1 correspondence with maps

$$Hom_{A-alg}(Sym_A^*(E^*),A) \cong Hom_{A-mod}(E^*,A) \cong E^{**} \cong E$$

since $E^{**} \cong E$ is a canonical isomorphism. Hence the module of global sections of the map $\pi$ is isomorphic to $E$ as $A$-module. We also have "by construction" an isomorphism $H^0(S, \mathcal{E}) \cong E$. You should also do the above mentioned exercise in HH.