Recall for $(\mathcal{C}, \otimes, a, c)$ a symmetric monoidal category, we defined a notion of a dualizable object and a trace $\tr(f \vert X)$ for any endomorphism $f \colon X \to X$. We can also define $$ \dim X = \tr(\id_X \vert X) \in \End(1). $$ This isn’t an integer but an element of $\End(1)$, so for instance, for $\mathsf{Vect}_ k$ with $k$ of characteristic $p$, we will have $\dim k^p = 0$.
Facts about the trace#
Proposition 1. Let $X, Y$ be dualizable objects in a symmetric monoidal category $\mathcal{C}$.
- $\tr(f \vert X) = \tr(f^\vee \vert X^\vee)$ for $f \colon X \to X$.
- $\tr(f \circ g \vert Y) = \tr(g \circ f \vert X)$ for $f \colon X \to Y$ and $g \colon Y \to X$.
- $\tr(f \otimes g \vert X \otimes Y) = \tr(f \vert X) \tr(g \vert Y)$ for $f \colon X \to X$ and $g \colon Y \to Y$.
- If $\mathcal{C}$ is additive and $f_i \colon X_i \to X_i$ then $$ \tr\Bigl( \bigoplus_i f_i \vert \bigoplus_i X_i \Bigr) = \sum_{i}^{} \tr(f_i \vert X_i). $$
- If $f \colon X \otimes Y \to X \otimes Y$, we can define the relative trace $$ \tr(f \vert X \otimes Y \vert Y) \colon X \to X \otimes Y \otimes Y^\vee \xrightarrow{f \otimes \id} X \otimes Y \otimes Y^\vee \to X. $$ Then $$ \tr(f \vert X \otimes Y) = \tr(\tr(f \vert X \otimes Y \vert Y) \vert X). $$
Proof.
Let’s only do (2). This follows from the diagram $$ \begin{CD} 1 @>>> X \otimes X^\vee @. \br @VVV @V{f \otimes 1}VV @. \br Y \otimes Y^\vee @>{1 \otimes f^\vee}>> Y \otimes X^\vee @>{1 \otimes g^\vee}>> Y \otimes Y^\vee \br @. @V{g \otimes 1}VV @VVV \br @. X \otimes X^\vee @>>> 1 \end{CD} $$ commuting.
Let $\mathcal{C}$ be an additive rigid monoidal category. We define $$ \mathcal{NC} = \lbrace f \colon X \to Y : \tr(f \circ g) = 0 \text{ for all } g \colon Y \to X \rbrace. $$
Corollary 2. The class of morphisms $\mathcal{NC}$ is a tensor ideal, meaning that it is stable under composition with an arbitrary morphism, and also tensoring with an arbitrary morphism.
So we can also define $$ \bar{\mathcal{C}} = \mathcal{C}/\mathcal{NC} $$ which has the same objects but the morphisms are $$ \Hom_{\bar{\mathcal{C}}}(X, Y) = \Hom_\mathcal{C}(X, Y) / (\mathcal{NC} \cap \Hom_\mathcal{C}(X, Y)) $$ where the quotient is as abelian groups. This is again going to be an additive rigid symmetric monoidal category.
Definition 3. Let $\mathcal{C}$ and $\mathcal{D}$ be two (symmetric) monoidal categories. A (symmetric) monoidal functor is a pair $(F, \lambda)$ where
- $F \colon \mathcal{C} \to \mathcal{D}$ is a functor,
- $\lambda$ is a functorial isomorphism $F(X) \otimes F(Y) \cong F(X \otimes Y)$ for $X, Y \in \mathcal{C}$,
such that
- $\lambda$ is compatible with $a$ (and $c$),
- $\lambda$ sends a unit (hence all units) to a unit.
Lemma 4.
- Let $F \colon \mathcal{C} \to \mathcal{D}$ be monoidal. If $X^\vee$ is a right dual of $X \in \mathcal{C}$, then $F(X^\vee)$ is a right dual of $F(X)$.
- If $F \colon \mathcal{C} \to \mathcal{D}$ is symmetric monoidal, and $X \in \mathcal{C}$ is dualizable, then $$ \begin{CD} \End_\mathcal{C}(X) @>{F}>> \End_\mathcal{D}(F(X)) \br @V{\tr}VV @V{\tr}VV \br \End_\mathcal{C}(1_\mathcal{C}) @>{F}>> \End_\mathcal{D}(1_\mathcal{D}) \end{CD} $$ commutes.
Abelian rigid symmetric monoidal categories#
Theorem 5. Let $(\mathcal{C}, \otimes, a, c)$ be a rigid symmetric monoidal abelian category. Suppose we have a commutative diagram $$ \begin{CD} 0 @>>> X^\prime @>>> X @>>> X^{\prime\prime} @>>> 0 \br @. @V{f^\prime}VV @V{f}VV @V{f^{\prime\prime}}VV @. \br 0 @>>> X^\prime @>>> X @>>> X^{\prime\prime} @>>> 0. \end{CD} $$ Then $$ \tr(f \vert X) = \tr(f^\prime \vert X^\prime) + \tr(f^{\prime\prime} \vert X^{\prime\prime}). $$ In particular, we get a map $$ \tr \colon K_0(\mathcal{C}) \to \End_\mathcal{C}(1). $$
Note that $- \otimes X$ preserves short exact sequences, because it commutes with all limits and colimits.
Lemma 6. A dual of a short exact sequence is a short exact sequence.
Proof.
We test against $\Hom(Y \otimes -)$.
Let $\mathcal{C}$ be an abelian rigid symmetric monoidal category. Let $F^\bullet \mathcal{C}$ be the category of finite filtered objects of $\mathcal{C}$, $$ 0 = F^n X \hookrightarrow \dotsb \hookrightarrow F^{m+1} X \hookrightarrow F^m = F^{m+1} = \dotsb. $$ We can define a tensor product here by $$ F^h(F^\bullet X \otimes F^\bullet Y) = \sum_{i+j=h}^{} F^i X \otimes F^j Y. $$ Note that this is rigid and additive (although no longer abelian), with the dual being $$ F^i (F^\bullet X)^\vee = (X / F^{1-i} X)^\vee. $$ The unit is given by $$ \dotsb = F^{-1} = 0 \subseteq 1 = F^0 = F^1 = \dotsb. $$
There are two functors $$ F^\bullet \mathcal{C} \to \mathcal{C}; \quad F^\bullet X \mapsto X $$ and also $$ F^\bullet \mathcal{C} \to \mathcal{C}^\mathbb{Z}; \quad F^\bullet X \mapsto \mathrm{gr}^\bullet X. $$ These are both symmetric monoidal, and our theorem about short exact sequences immediately follow from computing the trace after pushing forward $$ (0 \to X^\prime \to X = X) \in F^\bullet \mathcal{C} $$ in the two different directions.
Corollary 7. Let $\mathcal{C}$ be an abelian rigid symmetric monoidal category. If $f \colon X \to X$ is nilpotent.
Proof.
Filter $X$ by the kernels of the powers of $X$.
Let $\mathcal{C}$ be as above. If $U \hookrightarrow 1$ is a subobject, it turns out that $1$ splits as $$ 1 = U \oplus (1/U)^\vee. $$ Using this we can split $X$ into a direct sum of something over $U$ and another thing over $(1/U)^\vee$. In practice, we will just assume that the ring $k = \End(1)$ is a field, and so $1$ is irreducible.
Proposition 8. Let $\mathcal{C}$ be a pseudo-abelian (additive and idempotent-complete) rigid symmetric monoidal category. Suppose $\End_\mathcal{C} 1 = k$ is a field.
- If $\mathcal{C}$ is a semisimple abelian category, then $\mathcal{NC} = 0$.
- Suppose $\mathcal{NC} = 0$, $\tr(f) = 0$ for $f$ nilpotent, and $\Hom(X, Y)$ is finite-dimensional over $k$ for all $X, Y \in \mathcal{C}$. Then $\mathcal{C}$ is semisimple abelian.
Proof.
For (1), suppose $f \colon X \to Y$ is in $\mathcal{NC}$. We can break up $X$ and $Y$ into a direct sum of simple object, and then write it as a matrix $(f_{ij})$. We see that $f$ is in $\mathcal{NC}$ if and only if $(f_{ij})$ is in $\mathcal{NC}$ for all $f_{ij}$. We can only worry about $X_i = Y_j$, and so we are reduce to the case when $X = Y$ is irreducible.
We know that $\End(X)$ is a division algebra, but $\mathcal{NC} \cap \End(X_i)$ is a two-sided algebra, so if this is not zero, it must be everything. Now consider the $k$-linear map $$ \varphi \colon \Hom(1, X \otimes X^vee) \to \Hom(1, 1) = k $$ defined by composing with evaluation. The left hand side is a $\End(X)$-module of rank $1$, by using the action on $X$. This has the property that $\varphi(du) = 0$ for every $d \in \End(X)$, where $u$ is the unit map. On the other hand, the map $\varphi$ splits because we are in a semisimple category.
For (2), first note that $\End(X)$ is a semisimple $k$-algebra, because we are assuming that $\mathcal{C}$ is idempotent-complete. Consider $J_X \subseteq \End(X)$ the Jacobson radical. Then any element of $J_X$ is is nilpotent, because we’re inside a finite-dimensional algebra, and by assumption, these things have trace $0$. This implies that $J_X \subseteq \mathcal{NC} = 0$.
Example: fake motives#
Let $F$ be a field, and consider the category $\mathsf{SmProj}_ {/F}$ of smooth projective varieties over $F$. Given an adequate equivalence relation $\sim$ on the group $Z^\ast(X)_ k$ of cycles (with coefficients in a field $k$), one can define $$ A^\ast(X)_ k = Z^\ast(X)_ k / \sim $$ with a ring structure given by intersection product. There is also a well-defined pullback and pushforward map. There is also a composition map $$ A^\ast(X \times Y) \times A^\ast(Y \otimes Z) \to A^\ast(X \otimes Z) $$ and so $A^\ast(X \times X)$ as a ring structure.
Now one can define a category of fake motives by $$ \mathsf{Mot}_ F^\sim = \lbrace (X, p, m) \rbrace $$ where $X \in \mathsf{SmProj}_ {/F}$ and $p \in A^{\dim X}(X \times X)$ with $p^2 = p$ and $m \in \mathbb{Z}$. We then define morphisms as $$ \Hom((X, p, m), (Y, q, n)) = q A^{\dim X - m + n}(X \times Y) p. $$
Proposition 9. The category $\mathsf{Mot}_ F^\sim$ is pseudo-abelian rigid symmetric monoidal category, with $\End(1) = k$.
It turns out that $$ (X, \id, 0)^\vee = (X, \id, \dim X), $$ and for a self-correspondence $X \leftarrow Z \to X$, we have $$ \tr(Z \vert X) = \deg(Z \cap \Delta_X). $$ In this category, $\mathcal{NC}$ is precisely the numerical equivalence.
Corollary 10. $\mathsf{Mot}_ F^\mathrm{num}$ is semisimple abelian.