symmetric monoidal (∞,1)-category of spectra
In a field of positive characteristic, the usual derivative of polynomials has bad properties. Let $\mathbb{K}$ be such a field of characteristic $p \gt 0$. Consider the polynomial algebra $\mathbb{K}[X]$. The usual derivative of polynomials in one indeterminate defines a derivation $\mathbb{K}[X] \rightarrow \mathbb{K}[X]$ which associates $P'$ to $P$.
We then have that $(X^{p})'=p.X^{p-1}=0$ but $X^{p} \neq 0$ because $1^{p} = 1 \neq 0$. We thus lack the property that $P' = 0$ iff $P$ is a constant.
The notion of Hasse-Schmidt derivative comes as a replacement of the usual derivative and enjoys better properties.
It seems to be very linked to differential linear logic and to the notion of graded bimonoid.
Suppose that $R$ is a commutative rig. For every polynomial $P \in R[X]$ and $n \ge 0$, we define the $n-th$ Hasse-Schmidt derivative $P^{(n)}$ of $P$ by:
and if $n \ge 1$:
Suppose that $R$ is a commutative rig. We look at the polynomials in $R[X_{1},...,X_{q}]$. For every multiset $Y_{1}...Y_{n} \in \{X_{1},...,X_{q}\}$ and $P \in R[X_{1},...,X_{q}]$ we define the derivative $D_{Y_{1}...Y_{n}}(P)$. Note that we treat in the same way all the higher-order derivatives and define them directly and not by induction. It is in this way directly clear that the order of the variables with respect to we derivate doesn’t matter.
Suppose that $P$ is a monic monomial, that is a multiset of variables in $\{X_{1},...,X_{q}\}$.
If $Y_{1}...Y_{n}$ doesn’t divide $P$, then:
If $Y_{1}...Y_{n}$ divides $P$, then:
We then prolongate by linearity:
There is a very combinatorial interpretation which makes the link with the notion of graded bimonoid: if $P$ is a monic monomial, the derivative with respect to $Y_{1}...Y_{n}$ counts the number of ways to extract $Y_{1}...Y_{n}$ from $P$ and then multiplies it by $P$ where $Y_{1}...Y_{n}$ has been extracted. All the instances of a variable $Y_{i}$ in $P$ must be considered as different entities in this counting.
For example there is $\binom{3}{2}\binom{4}{1}$ ways to extract $X^{2}Y$ from $X^{3}Y^{4}$, we have $\binom{3}{2}$ choices of two instance of $X$ in $X^{3}$ and $\binom{4}{1}$ choices of two instances of $Y$ in $Y^{4}$, thus:
The notion was introduced in:
The definition and basic properties are reviewed on this blog page:
Last revised on August 18, 2022 at 18:35:50. See the history of this page for a list of all contributions to it.