I have recently been working on a project that involves tensor calculus. Here I collect some notions and identities, to spare myself from googling them each time.

We work on quasicoherent sheaves on a scheme, and when no ambiguity arises we omit this from the notation.

Schur functors

Here I should give an explanation of what Schur functors are and of how they generalize the symmetric and exterior products.

Instead, I refer the reader to chapter 6 of [2].

¯\_(ツ)_/¯

Identities

n(AB)    p+q=np(A)q(B)n(AB)    a=nΣa(A)Σa(B)Symn(AB)    p+q=nSymp(A)Symq(B)Symn(AB)    a=nΣa(A)Σa(B)\newcommand{\Sym}{\mathrm{Sym}} \begin{align} \bigwedge^{n}\left(A \oplus B\right) ~~&\simeq~~ \bigoplus_{p + q = n} \bigwedge^{p}(A) \otimes \bigwedge^{q}(B) \\ \bigwedge^{n}\left(A \otimes B\right) ~~&\simeq~~ \bigoplus_{|a| = n} \Sigma_{a} (A) \otimes \Sigma_{a^{*}} (B) \\ \Sym^{n}\left(A \oplus B\right) ~~&\simeq~~ \bigoplus_{p + q = n} \Sym^{p}(A) \otimes \Sym^{q}(B) \\ \Sym^{n}\left(A \otimes B\right) ~~&\simeq~~ \bigoplus_{|a| = n} \Sigma_{a} (A) \otimes \Sigma_{a} (B) \end{align}

Proof. The identities splitting direct sums are an immediate corollary of [Exercise II.5.16, 1].

The third item of the exercise states that for any short exact sequence 0ABC0,0 \to A \to B \to C \to 0, there are filtrations

0=Fn+1FnF1F0=Symn(B)\newcommand{\Sym}{\mathrm{Sym}} 0 = F_{n+1} \subset F_{n} \subset \dots \subset F_{1} \subset F_{0} = \Sym^{n}(B)

and

0=Gn+1GnG1G0=n(B),0 = G_{n+1} \subset G_{n} \subset \dots \subset G_{1} \subset G_{0} = \bigwedge^{n}(B),

such that

Fi/Fi+1Symi(A)Symni(C)\newcommand{\Sym}{\mathrm{Sym}} F_{i}/F_{i+1} \simeq \Sym^{i}(A) \otimes \Sym^{n-i}(C)

and

Gi/Gi+1i(A)ni(C).G_{i}/G_{i+1} \simeq \bigwedge^{i}(A) \otimes \bigwedge^{n-i}(C).

In the cases of two summands the proof is immediate, and the general cases follow by induction.

The identities splitting tensor products are proved in [Exercise 6.11, 2].

References

[1] Algebraic Geometry, R. Hartshorne.

[2] Representation Theory: A First Course, W. Fulton, J. Harris.