# nLab chain complex in super vector spaces

Contents

supersymmetry

## Applications

#### Homological algebra

homological algebra

Introduction

diagram chasing

# Contents

## Idea

A chain complex $V_\bullet$ of super vector spaces is for each $n \in \mathbb{Z}$ a super vector space $V_n = (V_n)_\even \oplus (V_n)_odd$, equipped with a differential, hence for each $n \in \mathbb{Z}$ a morphism $\partial_n \;\colon\; V_{n+1} \to V_n$ of super vector spaces such that $\partial \circ \partial = 0$.

The category $Ch(SuperVect)$ of chain complexes of super vector space is much like the category of chain complexes $Ch(Vect)$ of ordinary vector spaces, in particular in that it carries the tensor product of chain complexes that makes it a monoidal category. But there is a non-trivial symmetric braiding on $Ch(SuperVect)$ which involves not just the signs used in the braiding in $Ch(Vect)$, but also the signs involved in the defining nonp-trivial braiding on SuperVect. The commutative monoids with respect to this symmetric braiding on $Ch(SuperVect)$ are the differential graded-commutative superalgebras.

In fact there are two such such non-trivial symmetric braidings on $Ch(SuperVect)$ (Prop. below), but they are equivalent to each other (Prop. below).

## Definition

###### Definition

(chain complexes of super vector spaces)

Write $Ch(SuperVect)$ for the category of chain complexes inside the category of super vector spaces.

Hence for $V_\bullet \in Ch(SuperVect)$ an object, for each $n \in \mathbb{Z}$ there is a super vector space

$V_n = (V_n)_{even} \oplus (V_n)_{odd} \;\in SuperVect$

where we write the elements of the group of order two as $\mathbb{Z}/2 =\{even, odd\}$, with $even$ being the neutral element.

Hence we may regard any $V_\bullet \in Ch(SuperVect)$ equivalently as a $\mathbb{Z} \times (\mathbb{Z}/2)$-graded vector space equipped with a differential of degree $(1,even)$. For $v \in V$ an element in definite (“homogeneous”) bi-degree, we denote this degree by

(1)$deg(v) \coloneqq (n_v, \sigma_v) \;\in\; \mathbb{Z} \times (\mathbb{Z}/2) \,.$

The category $Ch(SuperVect)$ becomes a monoidal category under the tensor product of chain complexes applied to the tensor product of super vector spaces. This means that for $V, W \in Ch(SuperVect)$, the differential on a homogeneously graded element $v \otimes w \in V \otimes W$ is

$\partial(v \otimes w) \;=\; (\partial v) \otimes w + (-1)^{ n_v } v \otimes \partial w \,.$

## Properties

### Symmetric monoidal structure

###### Proposition

(symmetric monoidal structure on category of chain complexes of super vector spaces)

The monoidal category of chain complexes of super vector spaces $(Ch(SuperVect), \otimes)$ from Def. becomes a symmetric monoidal with each of the following two braiding isomorphisms, defined on tensor products of elements in homogenous bi-degree (1) as follows:

1. $\tau_{Deligne} \;\colon\; v \otimes w \mapsto (-1)^{ (n_v n_w + \sigma_v \sigma_w) } w \otimes v$;

2. $\tau_{Bernst} \;\colon\; v \otimes w \mapsto (-1)^{ (n_v + \sigma_v) (n_w + \sigma_w) } w \otimes v$.

Here in the exponents we are using the canonical ring-structure on the integers $\mathbb{Z}$ and on the prime field $\mathbb{Z}/2 = \mathbb{F}_2$, the implicit ring homomorphism $\mathbb{Z} \to \mathbb{Z}/2$ and we understand that $(-1)^{even} = 1$ and $(-1)^{odd} = -1$.

###### Proof

Since the expressions for both sign factors are symmetric in $v$ and $w$ in both cases, it is clear that $\tau_{w,v} \circ \tau_{v,w} = id_{v \otimes w}$ in both cases. Hence if $\tau$ is indeed a braiding, then it is symmetric.

To see that $\tau$ is indeed a braiding in each case, we need to check the hexagon identities

$\array{ (V_1 \otimes V_2) \otimes V_3 &\overset{a_{V_1, V_2, V_3}}{\longrightarrow}& V_1 \otimes (V_2 \otimes V_3) &\overset{\tau_{V_1,V_2 \otimes V_3}}{\longrightarrow}& (V_2 \otimes V_3) \otimes V_1 \\ \Big\downarrow{}^{\tau_{V_1,V_2}\otimes Id} &&&& \Big\downarrow{}^{a_{V_2,V_3,V_1}} \\ (V_2 \otimes V_1) \otimes V_3 &\overset{a_{V_2,V_1,V_3}}{\longrightarrow}& V_2 \otimes (V_1 \otimes V_3) &\overset{Id \otimes \tau_{V_1,V_3}}{\longrightarrow}& V_2 \otimes (V_3 \otimes V_1) }$

and

$\array{ V_1 \otimes (V_2 \otimes V_3) &\overset{a^{-1}_{V_1,V_2,V_3}}{\longrightarrow}& (V_1 \otimes V_2) \otimes V_3 &\overset{\tau_{V_1 \otimes V_2, V_3}}{\longrightarrow}& V_3 \otimes (V_1 \otimes V_2) \\ \Big\downarrow{}^{id \otimes \tau_{V_2,V_3}} &&&& \Big\downarrow{}^{a^{-1}_{V_3,V_1,V_2}} \\ V_1 \otimes (V_3 \otimes V_2) &\overset{a^{-1}_{V_1,V_3,V_2}}{\longrightarrow}& (V_1 \otimes V_3) \otimes V_2 &\overset{\tau_{V_1,V_3} \otimes id}{\longrightarrow}& (V_3 \otimes V_1) \otimes V_2 } \,.$

Since $\tau$ differs only by multiplication by a sign from the standard symmetric braiding on the category of vector spaces, which does satisfy its hexagon identities, it just remains to check that these sign factors picked up in going both ways around these diagrams agree.

Hence for $\tau_{Deligne}$ the two hexagon identities are equivalent to the conditions

$(-1)^{ n_1 (n_2 + n_3) + \sigma_1( \sigma_2 + \sigma_3 ) } \;=\; (-1)^{ n_1 n_2 + \sigma_1 \sigma_2 + n_1 n_3 + \sigma_1 \sigma_3 }$

and

$(-1)^{ (n_1 + n_2) n_3 + (\sigma_1 + \sigma_2) \sigma_3 } \;=\; (-1)^{ n_1 n_3 + \sigma_1 \sigma_3 + n_2 n_3 + \sigma_2 \sigma_3 }$

while for $\tau_{Bernst}$ they are equivalent to the conditions

$(-1)^{ (n_1 + \sigma_1)( n_2 + \sigma_2 + n_3 + \sigma_3 ) } \;=\; (-1)^{ (n_1 + \sigma_1)(n_2 + \sigma_2) + (n_1 + \sigma_1)(n_3 + \sigma_3) }$

and

$(-1)^{ (n_1 + \sigma_1 + n_2 + \sigma_2)(n_3 + \sigma_3) } \;=\; (-1)^{ (n_1 + \sigma_1)(n_3 + \sigma_3) + (n_2 + \sigma_2)(n_3 + \sigma_3) }$

for all triples of bi-degrees $(n_i, \sigma_i) \in \mathbb{Z} \times (\mathbb{Z}/2)$.

In both cases this holds because already the relevant exponents are equal in each case, by the distributive law for multiplication and addition in $\mathbb{Z}/2 = \mathbb{F}_2$.

###### Remark

(parity)

For the symmetric braiding $\tau_{Bernst}$ in Prop. it makes sense to name the additive group homomorphism

$\array{ \mathbb{Z} \times (\mathbb{Z}/2) &\overset{ parity }{\longrightarrow}& \mathbb{Z}/2 \\ (n , \sigma) &\mapsto& \sigma + n \, mod \, 2 }$

This is called parity in parts of the literature (following AKSZ 95). With this notation the Bernstein braiding in Prop. is simply

• $\tau_{Bernst} \;\colon\; v \otimes w \mapsto (-1)^{ parity(v) parity(w) } w \otimes v$.

Notice that for the other symmetric braiding $\tau_{Deligne}$, the concept of parity does not play any intrinsic role.

###### Proposition

(the two symmetric monoidal structures on the category of chain complexes of super vector spaces are equivalent)

The two symmetric monoidal category structures $\tau_{Deligne}$ and $\tau_{Bernst}$ on the monoidal category of chain complexes of super vector spaces $(Ch(SuperVect), \otimes)$ from Prop. are equivalent

$\tau_{Deligne} \;\simeq\; \tau_{Bernst}$

in that the identity functor $Ch(SuperVect) \overset{id}{\longrightarrow} Ch(SuperVect)$ equipped with the following monoidal natural isomorphism

$\array{ id(V) \otimes id(W) \overset{\mu}{\longrightarrow} id(V \otimes W) \\ v \otimes w \;\mapsto\; (-1)^{ n_v \sigma_w } \, v \otimes w }$

(the second line shows its action on elements of homogeneous bidegree $(n,\sigma) \in \mathbb{Z} \times (\mathbb{Z}/2)$)

becomes a strong symmetric monoidal functor

$(Ch(SuperVect), \otimes, \tau_{Deligne}) \underoverset{\simeq}{(id,\mu)}{\longrightarrow} (Ch(SuperVect), \otimes, \tau_{Bernst}) \,.$
###### Proof

First to see that we have a strong monoidal functor, we need to check the associativity condition

$\array{ (id(V_1) \otimes id(V_2)) \otimes id(V_3) &\underoverset{\simeq}{a_{id(V_1),id(V_2),id(V_3)}}{\longrightarrow}& id(V_1) \otimes( id(V_2)\otimes id(V_3) ) \\ {}^{\mathllap{\mu_{V_1,V_2} \otimes id}}\downarrow && \Big\downarrow{}^{\mathrlap{id\otimes \mu_{V_2,V_3}}} \\ id(V_1 \otimes V_2) \otimes id(V_3) && id(V_1) \otimes id(V_2 \otimes V_3) \\ {}^{\mathllap{\mu_{V_1 \otimes V_2 , V_3} } }\Big\downarrow && \Big\downarrow^{\mathrlap{\mu_{ V_1, V_2 \otimes V_3 }}} \\ id( ( V_1 \otimes V_2 ) \otimes V_3 ) &\underset{id(a_{V_1,V_2,V_3})}{\longrightarrow}& id( V_1 \otimes ( V_2 \otimes V_3 ) ) } \,,$

and the unitality conditions

$\array{ 1 \otimes id(V) &\overset{\epsilon \otimes id}{\longrightarrow}& id(1) \otimes id(V) \\ {}^{\mathllap{\ell_{id(V)}}}\Big\downarrow && \Big\downarrow{}^{\mathrlap{\mu_{1, V }}} \\ id(V) &\overset{id(\ell_V )}{\longleftarrow}& id(1 \otimes V ) }$

and

$\array{ id(V) \otimes 1 &\overset{id \otimes \epsilon }{\longrightarrow}& id(V) \otimes id(1) \\ {}^{\mathllap{r_{id(V)}}}\Big\downarrow && \Big\downarrow{}^{\mathrlap{\mu_{V, 1 }}} \\ id(V) &\overset{F(r_V )}{\longleftarrow}& id(V \otimes 1 ) } \,,$

Since $\mu$ differs from the trivial monoidal isomorphism only by the sign factor, this is equivalent to the condition that the sign factors picked up in going both ways around these diagrams agree.

For associativity this is the condition

$(-1)^{ n_1 \sigma_2 + (n_1 + n_2)\sigma_3 } \;=\; (-1)^{ n_2 \sigma_3 + n_1 (\sigma_2 + \sigma_3) }$

for all bi-degrees $(n_i, \sigma_i) \in \mathbb{Z} \times (\mathbb{Z}/2)$, which holds, because it already holds for the exponents themselves, as an identity in $\mathbb{Z}/2 =\mathbb{F}_2$.

For the unitality condition this is the statement that the sign given by $\mu_{1,x}$ and $\mu_{x,1}$ is the the trivial sign $(-1)^0 = 1$. This is indeed the case because the tensor unit is in degree $0 = (0,even) \in \mathbb{Z} \times (\mathbb{Z}/2)$.

Now to see that we have a symmetric monoidal functor, we need to show that it intertwines the two symmetruc braiding isomorphisms

$\array{ id(V_1) \otimes id(V_2) &\stackrel{\tau_{Bernst}}{\longrightarrow}& id(V_2) \otimes id(V_1) \\ {}^{\mathllap{\mu_{V_1,V_2}}}\Big\downarrow && \Big\downarrow{}^{\mathrlap{\mu_{V_2,V_1}}} \\ id(V_1\otimes V_2) &\overset{id(\tau_{Deligne})}{\longrightarrow}& if(W \otimes V) }$

As before, this is equivalent to a condition on the signs picked up both ways, which reads:

$(-1)^{ (n_1 + \sigma_1)(n_2 + \sigma_2) + n_2 \sigma_1} \;=\; (-1)^{ n_1 \sigma_2 + n_1 n_2 + \sigma_1 \sigma_2 } \,.$

Inspection shows that this is indeed the case:

The two signs of $\tau_{Deligne}$ and $\tau_{Bernst}$ differ by the “mixed terms” that are produced in multiplying out $(n_v + \sigma_v)(n_w + \sigma_w)$ and these two mixed terms is just what the two occurences $\mu$ provides (using that $+1 = -1 \in \mathbb{F}_2$).

### Differential graded-commutative superalgebras

###### Definition

A differential graded-commutative superalgebra is a commutative monoid in the symmetric monoidal category of chain complexes of super vector spaces, for either of the symmetric monoidal structures from Prop. .

$sdgcAlg_{Deligne} \;\coloneqq\; ComMon\left( Ch(SuperVect), \otimes, \tau_{Deligne} \right)$
$sdgcAlg_{Benrst} \;\coloneqq\; ComMon\left( Ch(SuperVect), \otimes, \tau_{Bernst} \right)$

The resulting sign rule is this:

sign rule for differential graded-commutative superalgebras
(different but equivalent)

$\phantom{A}$Deligne’s convention$\phantom{A}$$\phantom{A}$Bernstein’s convention$\phantom{A}$
$\phantom{A}$$\alpha_i \cdot \alpha_j =$$\phantom{A}$$\phantom{A}$$(-1)^{ (n_i \cdot n_j + \sigma_i \cdot \sigma_j) } \alpha_j \cdot \alpha_i$$\phantom{A}$$\phantom{A}$$(-1)^{ (n_i + \sigma_i) \cdot (n_j + \sigma_j) } \alpha_j \cdot \alpha_i$$\phantom{A}$
$\phantom{A}$common in$\phantom{A}$
$\phantom{A}$discussion of$\phantom{A}$
$\phantom{A}$supergravity$\phantom{A}$$\phantom{A}$AKSZ sigma-models$\phantom{A}$
$\phantom{A}$representative$\phantom{A}$
$\phantom{A}$references$\phantom{A}$
$\phantom{A}$Bonora et. al 87,$\phantom{A}$
$\phantom{A}$Castellani-D’Auria-Fré 91,$\phantom{A}$
$\phantom{A}$Deligne-Freed 99$\phantom{A}$
$\phantom{A}$AKSZ 95,$\phantom{A}$
$\phantom{A}$Carchedi-Roytenberg 12$\phantom{A}$

Since the identity functor carries a strong symmetric monoidal equivalence of categories (Prop. )

$\left( Ch(SuperVect), \otimes, \tau_{Deligne} \right) \;\simeq\; \left( Ch(SuperVect), \otimes, \tau_{Bernst} \right)$

there is (via this Prop.) an induced equivalence of categories of the corresponding symmetric monoids

$ComMon\left( Ch(SuperVect), \otimes, \tau_{Deligne} \right) \;\simeq\; ComMon\left( Ch(SuperVect), \otimes, \tau_{Bernst} \right)$

hence of these two versions of differential graded-commutative superalgebras:

$sdgcAlg_{Deligne} \;\simeq\; sdgcAlg_{Bernst}$