nLab I.10, Frobenius morphism and symmetric products

This entry is about a section of the text

• Michel Demazure, lectures on p-divisible groups, web

We give here another characterization of the Frobenius morphism in terms of symmetric products.

Let $p$ be a prime number, let $k$ be a filed of characteristic $p$, let $V$ be a $k$-vector space, let $\otimes^p V$ denote the $p$-fold tensor power of $V$, let $TS^p V$ denote the subspace of symmetric tensors. Then we have the symmetrization operator

$$s_V: \begin{cases} \otimes^p V\to TS^p V \\ a_1\otimes\cdots\otimes a_n\mapsto \Sigma_{\sigma\in S_p}a_{\sigma(1)}\otimes\cdots\otimes a_{\sigma(n)} \end{cases}$$

end the linear map

$$\alpha_V: \begin{cases} TS^p V\to\otimes^p V \\ a\otimes \lambda\mapsto\lambda(a\otimes\cdots\otimes a) \end{cases}$$

then the map $V^{(p)}\stackrel{\alpha_V}{\to}TS^p V\to TS^p V/s(\otimes^p V)$ is bijective and we define $\lambda_V:TS^p V\to V^{(p)}$ by

$$\lambda_V\circ s=0$$

and

$$\lambda_V \circ \alpha_V= id$$

If $A$ is a $k$-ring we have that $TS^p A$ is a $k$-ring and $\lambda_A$ is a $k$-ring morphism.

If $X=Sp_k A$ is a ring spectrum we abbreviate $S^p X=S^p_k X:=Sp_k (TS^p A)$ and the following diagram is commutative.

$$\array{ X &\stackrel{F_X}{\to}& X^{(p)} \\ \downarrow&&\downarrow \\ X^p &\stackrel{can}{\to}& S^p X }$$