## 12.17 Additive monoidal categories

Some material about the interaction between a monoidal structure and an additive structure on a category.

Definition 12.17.1. An additive monoidal category is an additive category $\mathcal{A}$ endowed with a monoidal structure $\otimes , \phi$ (Categories, Definition 4.42.1) such that $\otimes$ is an additive functor in each variable.

Lemma 12.17.2. Let $\mathcal{A}$ be an additive monoidal category. If $Y_ i$, $i = 1, 2$ are left duals of $X_ i$, $i = 1, 2$, then $Y_1 \oplus Y_2$ is a left dual of $X_1 \oplus X_2$.

Proof. Follows from uniqueness of adjoints and Categories, Remark 4.42.7. $\square$

Lemma 12.17.3. In a Karoubian additive monoidal category every summand of an object which has a left dual has a left dual.

Proof. We will use Categories, Lemma 4.42.6 without further mention. Let $X$ be an object which has a left dual $Y$. We have

$\mathop{\mathrm{Hom}}\nolimits (X, X) = \mathop{\mathrm{Hom}}\nolimits (\mathbf{1}, X \otimes Y) = \mathop{\mathrm{Hom}}\nolimits (Y, Y)$

If $a : X \to X$ corresponds to $b : Y \to Y$ then $b$ is the unique endomorphism of $Y$ such that precomposing by $a$ on

$\mathop{\mathrm{Hom}}\nolimits (Z' \otimes X, Z) = \mathop{\mathrm{Hom}}\nolimits (Z', Z \otimes Y)$

is the same as postcomposing by $1 \otimes b$. Hence the bijection $\mathop{\mathrm{Hom}}\nolimits (X, X) \to \mathop{\mathrm{Hom}}\nolimits (Y, Y)$, $a \mapsto b$ is an isomorphism of the opposite of the algebra $\mathop{\mathrm{Hom}}\nolimits (X, X)$ with the algebra $\mathop{\mathrm{Hom}}\nolimits (Y, Y)$. In particular, if $X = X_1 \oplus X_2$, then the corresponding projectors $e_1, e_2$ are mapped to idempotents in $\mathop{\mathrm{Hom}}\nolimits (Y, Y)$. If $Y = Y_1 \oplus Y_2$ is the corresponding direct sum decomposition of $Y$ (Section 12.4) then we see that under the bijection $\mathop{\mathrm{Hom}}\nolimits (Z' \otimes X, Z) = \mathop{\mathrm{Hom}}\nolimits (Z', Z \otimes Y)$ we have $\mathop{\mathrm{Hom}}\nolimits (Z' \otimes X_ i, Z) = \mathop{\mathrm{Hom}}\nolimits (Z', Z \otimes Y_ i)$ functorially as subgroups for $i = 1, 2$. It follows that $Y_ i$ is the left dual of $X_ i$ by the discussion in Categories, Remark 4.42.7. $\square$

Example 12.17.4. Let $F$ be a field. Let $\mathcal{C}$ be the category of graded $F$-vector spaces. Given graded vector spaces $V$ and $W$ we let $V \otimes W$ denote the graded $F$-vector space whose degree $n$ part is

$(V \otimes W)^ n = \bigoplus \nolimits _{n = p + q} V^ p \otimes _ F W^ q$

Given a third graded vector space $U$ as associativity constraint $\phi : U \otimes (V \otimes W) \to (U \otimes V) \otimes W$ we use the “usual” isomorphisms

$U^ p \otimes _ F (V^ q \otimes _ F W^ r) \to (U^ p \otimes _ F V^ q) \otimes _ F W^ r$

of vectors spaces. As unit we use the graded $F$-vector space $\mathbf{1}$ which has $F$ in degree $0$ and is zero in other degrees. There are two commutativity constraints on $\mathcal{C}$ which turn $\mathcal{C}$ into a symmetric monoidal category: one involves the intervention of signs and the other does not. We will usually use the one that does. To be explicit, if $V$ and $W$ are graded $F$-vector spaces we will use the isomorphism $\psi : V \otimes W \to W \otimes V$ which in degree $n$ uses

$V^ p \otimes _ F W^ q \to W^ q \otimes _ F V^ p,\quad v \otimes w \mapsto (-1)^{pq} w \otimes v$

We omit the verification that this works.

Lemma 12.17.5. Let $F$ be a field. Let $\mathcal{C}$ be the category of graded $F$-vector spaces viewed as a monoidal category as in Example 12.17.4. If $V$ in $\mathcal{C}$ has a left dual $W$, then $\sum _ n \dim _ F V^ n < \infty$ and the map $\epsilon$ defines nondegenerate pairings $W^{-n} \times V^ n \to F$.

Proof. As unit we take By Categories, Definition 4.42.5 we have maps

$\eta : \mathbf{1} \to V \otimes W\quad \epsilon : W \otimes V \to \mathbf{1}$

Since $\mathbf{1} = F$ placed in degree $0$, we may think of $\epsilon$ as a sequence of pairings $W^{-n} \times V^ n \to F$ as in the statement of the lemma. Choose bases $\{ e_{n, i}\} _{i \in I_ n}$ for $V^ n$ for all $n$. Write

$\eta (1) = \sum e_{n, i} \otimes w_{-n, i}$

for some elements $w_{-n, i} \in W^{-n}$ almost all of which are zero! The condition that $(\epsilon \otimes 1) \circ (1 \otimes \eta )$ is the identity on $W$ means that

$\sum \nolimits _{n, i} \epsilon (w, e_{n, i})w_{-n, i} = w$

Thus we see that $W$ is generated as a graded vector space by the finitely many nonzero vectors $w_{-n, i}$. The condition that $(1 \otimes \epsilon ) \circ (\eta \otimes 1)$ is the identity of $V$ means that

$\sum \nolimits _{n, i} e_{n, i}\ \epsilon (w_{-n, i}, v) = v$

In particular, setting $v = e_{n, i}$ we conclude that $\epsilon (w_{-n, i}, e_{n, i'}) = \delta _{ii'}$. Thus we find that the statement of the lemma holds and that $\{ w_{-n, i}\} _{i \in I_ n}$ is the dual basis for $W^{-n}$ to the chosen basis for $V^ n$. $\square$

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