Super-Preadditive Sketches and the Koszul Tensor Product1. Motivation2. Super abelian groups3. Super-preadditive sketches4. Forgetting parity5. Koszul tensor product of super-preadditive sketches6. Koszul interchange law7. Internal Hom8. Tensor--Hom adjunction9. Proof of the tensor--Hom adjunction10. Example: anticommuting double complexes

 

Super-Preadditive Sketches and the Koszul Tensor Product

1. Motivation

Ordinary preadditive sketches are small categories enriched over

$$\mathbf{Ab}.$$

Their tensor product gives commuting interchange laws. In particular, if $\mathsf{ChSk}$$\mathsf{ChSk}$ is the preadditive sketch for chain complexes, then

$$\mathsf{ChSk}\otimes \mathsf{ChSk}$$

gives a sketch for commuting bicomplexes:

$$d^v{d}^h=d^h{d}^v.$$

However, the double complexes used in homological algebra usually satisfy the anticommuting relation

$$d^v{d}^h+d^h{d}^v=0.$$

Equivalently,

$$d^v{d}^h=-d^h{d}^v.$$

This sign is not visible in the ordinary tensor product of preadditive sketches. To encode it structurally, we pass from $\mathbf{Ab}$$\mathbf{Ab}$-enrichment to super $\mathbf{Ab}$$\mathbf{Ab}$-enrichment.

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2. Super abelian groups

Let

$$\mathbf{SAb}$$

denote the category of super abelian groups, that is, $\mathbb{Z}/2$$\mathbb Z/2$-graded abelian groups

$$A=A_{\overline{0}}\oplus A_{\overline{1}}.$$

A homogeneous element $a\in A$$a\in A$ has degree

$$|a|\in \mathbb{Z}/2.$$

The tensor product is the usual graded tensor product:

$$(A\otimes B{)}_{\overline{r}}=\underset{\overline{p}+\overline{q}=\overline{r}}{⨁}{A}_{\overline{p}}\otimes B_{\overline{q}}.$$

The unit object is

$$\mathbb{Z}$$

concentrated in degree $\overline{0}$$\bar 0$.

The symmetric braiding is the Koszul braiding

$${\tau }_{A,B}:A\otimes B\to B\otimes A,$$

given on homogeneous elements by

$${\tau }_{A,B}(a\otimes b)=(-1{)}^{|a||b|}b\otimes a.$$

Thus $\mathbf{SAb}$$\mathbf{SAb}$ is a symmetric monoidal category.

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3. Super-preadditive sketches

A super-preadditive sketch is a small $\mathbf{SAb}$$\mathbf{SAb}$-enriched category.

Thus a super-preadditive sketch $\mathsf{S}$$\mathsf S$ consists of:

1. a set of objects;

2. for each pair $s,s^{\prime }$$s,s'$, a super abelian group

   $$\mathsf{S}(s,s^{\prime });$$

3. composition morphisms in $\mathbf{SAb}$$\mathbf{SAb}$

   $$\mathsf{S}(s^{\prime },s^{″})\otimes \mathsf{S}(s,s^{\prime })⟶\mathsf{S}(s,s^{″});$$

4. unit morphisms

   $$\mathbb{Z}\to \mathsf{S}(s,s).$$

Composition is degree-preserving. Thus if $f$$f$ and $g$$g$ are homogeneous and composable, then

$$|g\circ f|=|g|+|f|.$$

The identity morphisms are even:

$$|1_s|=0.$$

As before, we are still considering cone-free sketches. The new feature is not the presence of cones or cocones, but the presence of parity.

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4. Forgetting parity

There is a forgetful operation

$$\mathrm{Und}:\mathbf{SAb}\text{-}\mathbf{Cat}\to \mathbf{Ab}\text{-}\mathbf{Cat}.$$

It sends a super-preadditive sketch $\mathsf{S}$$\mathsf S$ to the ordinary preadditive sketch $\mathrm{Und}(\mathsf{S})$$\operatorname{Und}(\mathsf S)$ with the same objects and Hom groups

$$\mathrm{Und}(\mathsf{S})(s,s^{\prime })=\mathsf{S}(s,s^{\prime }{)}_{\overline{0}}\oplus \mathsf{S}(s,s^{\prime }{)}_{\overline{1}}.$$

This point matters.

If $\mathcal{A}$$\mathcal A$ is an ordinary additive category, we usually do not view it as a super category concentrated in even degree when interpreting chain-complex sketches with odd differentials. If we did that, every odd generator would have to map to zero.

Instead, for ordinary additive semantics, we use

$$\mathrm{Und}(\mathsf{S})$$

and define an $\mathsf{S}$$\mathsf S$-model in $\mathcal{A}$$\mathcal A$ as an additive functor

$$\mathrm{Und}(\mathsf{S})\to \mathcal{A}.$$

Thus parity is a syntactic device used to construct the correct signed relations. After the signed sketch has been constructed, one may forget parity and take ordinary additive models.

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5. Koszul tensor product of super-preadditive sketches

Let $\mathsf{S}$$\mathsf S$ and $\mathsf{T}$$\mathsf T$ be super-preadditive sketches. Their Koszul tensor product

$$\mathsf{S}\hat{\otimes }\mathsf{T}$$

is defined as follows.

The objects are pairs

$$(s,t).$$

The Hom supergroup is

$$(\mathsf{S}\hat{\otimes }\mathsf{T})((s,t),(s^{\prime },t^{\prime }))=\mathsf{S}(s,s^{\prime })\otimes \mathsf{T}(t,t^{\prime }).$$

The composition is defined using the Koszul braiding. Structurally, it is the composite

$$\begin{array}{rl}& (\mathsf{S}(s^{\prime },s^{″})\otimes \mathsf{T}(t^{\prime },t^{″}))\otimes (\mathsf{S}(s,s^{\prime })\otimes \mathsf{T}(t,t^{\prime }))\\ & ⟶(\mathsf{S}(s^{\prime },s^{″})\otimes \mathsf{S}(s,s^{\prime }))\otimes (\mathsf{T}(t^{\prime },t^{″})\otimes \mathsf{T}(t,t^{\prime }))\\ & ⟶\mathsf{S}(s,s^{″})\otimes \mathsf{T}(t,t^{″}).\end{array}$$

The first arrow uses the Koszul braiding to move

$$\mathsf{T}(t^{\prime },t^{″})$$

past

$$\mathsf{S}(s,s^{\prime }).$$

In element notation, for homogeneous morphisms,

$$(f_2\otimes g_2)\circ (f_1\otimes g_1)=(-1{)}^{|g_2||f_1|}(f_2{f}_1)\otimes (g_2{g}_1).$$

The identity on $(s,t)$$(s,t)$ is

$$1_{(s,t)}=1_s\otimes 1_t.$$

When all morphisms are even, the sign disappears and this reduces to the ordinary tensor product of preadditive sketches.

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6. Koszul interchange law

Let

$$f:s\to s^{\prime }$$

be a homogeneous morphism in $\mathsf{S}$$\mathsf S$, and let

$$g:t\to t^{\prime }$$

be a homogeneous morphism in $\mathsf{T}$$\mathsf T$.

Inside

$$\mathsf{S}\hat{\otimes }\mathsf{T},$$

we have the two morphisms

$$f\otimes 1_t:(s,t)\to (s^{\prime },t)$$

and

$$1_s\otimes g:(s,t)\to (s,t^{\prime }).$$

Compare the two composites around the square

$$\begin{array}{ccc}(s,t)& \stackrel{f\otimes 1_t}{\to }& (s^{\prime },t)\\ \downarrow 1_s\otimes g& & \downarrow 1_{s^{\prime }}\otimes g\\ (s,t^{\prime })& \stackrel{f\otimes 1_{t^{\prime }}}{\to }& (s^{\prime },t^{\prime }).\end{array}$$

One path gives

$$(1_{s^{\prime }}\otimes g)\circ (f\otimes 1_t)=(-1{)}^{|f||g|}(f\otimes g).$$

The other path gives

$$(f\otimes 1_{t^{\prime }})\circ (1_s\otimes g)=f\otimes g.$$

Therefore

$$(1_{s^{\prime }}\otimes g)\circ (f\otimes 1_t)=(-1{)}^{|f||g|}(f\otimes 1_{t^{\prime }})\circ (1_s\otimes g).$$

This is the Koszul interchange law.

If $f$$f$ and $g$$g$ are both odd, then

$$(-1{)}^{|f||g|}=-1,$$

and the square anticommutes.

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7. Internal Hom

Let $\mathsf{T}$$\mathsf T$ and $\mathsf{U}$$\mathsf U$ be super-preadditive sketches. Define

$$[\mathsf{T},\mathsf{U}]$$

as follows.

The objects are even $\mathbf{SAb}$$\mathbf{SAb}$-enriched functors

$$F:\mathsf{T}\to \mathsf{U}.$$

For two such functors $F,G$$F,G$, the Hom supergroup is the super abelian group of graded natural transformations

$$F\Rightarrow G.$$

Equivalently, it is the enriched end

$$[\mathsf{T},\mathsf{U}](F,G)={\int }_{t\in \mathsf{T}}\mathsf{U}(Ft,Gt).$$

Concretely, a homogeneous element

$$\eta \in [\mathsf{T},\mathsf{U}](F,G)$$

of degree $r\in \mathbb{Z}/2$$r\in \mathbb Z/2$ is a family of homogeneous morphisms

$${\eta }_t:Ft\to Gt$$

of degree $r$$r$, satisfying the graded naturality condition

$$G(a)\circ {\eta }_t=(-1{)}^{r|a|}{\eta }_{t^{\prime }}\circ F(a)$$

for every homogeneous morphism

$$a:t\to t^{\prime }$$

in $\mathsf{T}$$\mathsf T$.

Addition is pointwise, and composition is pointwise:

$$(\theta \circ \eta {)}_t={\theta }_t\circ {\eta }_t.$$

This makes

$$[\mathsf{T},\mathsf{U}]$$

into a super-preadditive sketch.

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8. Tensor--Hom adjunction

The Koszul tensor product satisfies the tensor--Hom adjunction

$$\boxed{{\displaystyle \mathrm{SPreAdd}(\mathsf{S}\hat{\otimes }\mathsf{T},\mathsf{U})\cong \mathrm{SPreAdd}(\mathsf{S},[\mathsf{T},\mathsf{U}]).}}$$

Here $\mathrm{SPreAdd}$$\operatorname{SPreAdd}$ denotes the category of small super-preadditive sketches and even enriched functors.

Thus the category of small super-preadditive sketches is monoidal closed.

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9. Proof of the tensor--Hom adjunction

Let

$$H:\mathsf{S}\hat{\otimes }\mathsf{T}\to \mathsf{U}$$

be an even enriched functor.

We construct

$$\tilde{H}:\mathsf{S}\to [\mathsf{T},\mathsf{U}].$$

For each object $s\in \mathsf{S}$$s\in\mathsf S$, define

$$\tilde{H}(s):\mathsf{T}\to \mathsf{U}$$

by

$$\tilde{H}(s)(t)=H(s,t),$$

and

$$\tilde{H}(s)(g)=H(1_s\otimes g).$$

Now let

$$f:s\to s^{\prime }$$

be a homogeneous morphism in $\mathsf{S}$$\mathsf S$. Define a homogeneous natural transformation

$$\tilde{H}(f):\tilde{H}(s)\Rightarrow \tilde{H}(s^{\prime })$$

of degree $|f|$$|f|$ by

$$\tilde{H}(f{)}_t=H(f\otimes 1_t).$$

We verify graded naturality. Let

$$g:t\to t^{\prime }$$

be homogeneous. We need

$$\tilde{H}(s^{\prime })(g)\circ \tilde{H}(f{)}_t=(-1{)}^{|f||g|}\tilde{H}(f{)}_{t^{\prime }}\circ \tilde{H}(s)(g).$$

The left-hand side is

$$H(1_{s^{\prime }}\otimes g)\circ H(f\otimes 1_t)=H((1_{s^{\prime }}\otimes g)\circ (f\otimes 1_t)).$$

By the Koszul interchange law,

$$(1_{s^{\prime }}\otimes g)\circ (f\otimes 1_t)=(-1{)}^{|f||g|}(f\otimes 1_{t^{\prime }})\circ (1_s\otimes g).$$

Therefore

$$\tilde{H}(s^{\prime })(g)\circ \tilde{H}(f{)}_t=(-1{)}^{|f||g|}H((f\otimes 1_{t^{\prime }})\circ (1_s\otimes g)).$$

This is

$$(-1{)}^{|f||g|}\tilde{H}(f{)}_{t^{\prime }}\circ \tilde{H}(s)(g).$$

So $\tilde{H}(f)$$\widetilde H(f)$ is a graded natural transformation.

Hence

$$\tilde{H}:\mathsf{S}\to [\mathsf{T},\mathsf{U}]$$

is an even enriched functor.

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Conversely, suppose we have an even enriched functor

$$K:\mathsf{S}\to [\mathsf{T},\mathsf{U}].$$

We construct

$$\hat{K}:\mathsf{S}\hat{\otimes }\mathsf{T}\to \mathsf{U}.$$

On objects, define

$$\hat{K}(s,t)=K(s)(t).$$

Let

$$f:s\to s^{\prime }$$

and

$$g:t\to t^{\prime }$$

be homogeneous morphisms. Define

$$\hat{K}(f\otimes g)=K(f{)}_{t^{\prime }}\circ K(s)(g).$$

This is the “vertical first, then horizontal” formula.

By graded naturality of

$$K(f):K(s)\Rightarrow K(s^{\prime }),$$

we have

$$K(s^{\prime })(g)\circ K(f{)}_t=(-1{)}^{|f||g|}K(f{)}_{t^{\prime }}\circ K(s)(g).$$

Hence equivalently,

$$\hat{K}(f\otimes g)=(-1{)}^{|f||g|}K(s^{\prime })(g)\circ K(f{)}_t.$$

The construction is bilinear and degree-preserving, so it extends to all Hom supergroups.

Now check composition. Let

$$f_1:s_0\to s_1,f_2:s_1\to s_2$$

be homogeneous morphisms in $\mathsf{S}$$\mathsf S$, and let

$$g_1:t_0\to t_1,g_2:t_1\to t_2$$

be homogeneous morphisms in $\mathsf{T}$$\mathsf T$.

Then

$$(f_2\otimes g_2)\circ (f_1\otimes g_1)=(-1{)}^{|g_2||f_1|}(f_2{f}_1)\otimes (g_2{g}_1).$$

On the other hand,

$$\hat{K}(f_2\otimes g_2)\circ \hat{K}(f_1\otimes g_1)$$

equals

$$(K(f_2{)}_{t_2}\circ K(s_1)(g_2))\circ (K(f_1{)}_{t_1}\circ K(s_0)(g_1)).$$

Using graded naturality of $K(f_1)$$K(f_1)$, we have

$$K(s_1)(g_2)\circ K(f_1{)}_{t_1}=(-1{)}^{|f_1||g_2|}K(f_1{)}_{t_2}\circ K(s_0)(g_2).$$

Therefore

$$\begin{array}{rl}& \hat{K}(f_2\otimes g_2)\circ \hat{K}(f_1\otimes g_1)\\ & =(-1{)}^{|f_1||g_2|}K(f_2{)}_{t_2}\circ K(f_1{)}_{t_2}\circ K(s_0)(g_2)\circ K(s_0)(g_1)\\ & =(-1{)}^{|g_2||f_1|}K(f_2{f}_1{)}_{t_2}\circ K(s_0)(g_2{g}_1)\\ & =\hat{K}((-1{)}^{|g_2||f_1|}(f_2{f}_1)\otimes (g_2{g}_1))\\ & =\hat{K}((f_2\otimes g_2)\circ (f_1\otimes g_1)).\end{array}$$

Thus $\hat{K}$$\widehat K$ preserves composition. It also preserves identities, so it is an even enriched functor

$$\hat{K}:\mathsf{S}\hat{\otimes }\mathsf{T}\to \mathsf{U}.$$

The two constructions

$$H\mapsto \tilde{H}$$

and

$$K\mapsto \hat{K}$$

are inverse to each other, by direct inspection on objects and elementary tensors. Therefore

$$\mathrm{SPreAdd}(\mathsf{S}\hat{\otimes }\mathsf{T},\mathsf{U})\cong \mathrm{SPreAdd}(\mathsf{S},[\mathsf{T},\mathsf{U}]).$$

This proves the tensor--Hom adjunction.

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10. Example: anticommuting double complexes

Let

$${\mathsf{ChSk}}^{\sup }$$

be the super-preadditive sketch generated by objects

$$n\in \mathbb{Z}$$

and odd arrows

$$d_n:n\to n-1$$

subject to

$$d_{n-1}{d}_n=0.$$

Now form

$${\mathsf{ChSk}}^{\sup }\hat{\otimes }{\mathsf{ChSk}}^{\sup }.$$

Its objects are pairs

$$(p,q).$$

It has horizontal differentials

$$d^h=d_p\otimes 1_q$$

and vertical differentials

$$d^v=1_p\otimes d_q.$$

First, going horizontally and then vertically gives

$$d^v{d}^h=(1_{p-1}\otimes d_q)\circ (d_p\otimes 1_q).$$

By the Koszul composition rule,

$$(f_2\otimes g_2)\circ (f_1\otimes g_1)=(-1{)}^{|g_2||f_1|}(f_2{f}_1)\otimes (g_2{g}_1).$$

Here

$$f_2=1_{p-1},g_2=d_q,f_1=d_p,g_1=1_q.$$

Since $d_p$$d_p$ and $d_q$$d_q$ are both odd, we get

$$(-1{)}^{|g_2||f_1|}=(-1{)}^{|d_q||d_p|}=(-1{)}^{1\cdot 1}=-1.$$

Therefore

$$(1_{p-1}\otimes d_q)\circ (d_p\otimes 1_q)=-(d_p\otimes d_q).$$

On the other hand, going vertically and then horizontally gives

$$d^h{d}^v=(d_p\otimes 1_{q-1})\circ (1_p\otimes d_q).$$

Again using the Koszul composition rule, the sign is now

$$(-1{)}^{|1_{q-1}||1_p|}=(-1{)}^{0\cdot 0}=1,$$

because identity morphisms are even. Hence

$$(d_p\otimes 1_{q-1})\circ (1_p\otimes d_q)=d_p\otimes d_q.$$

Thus

$$d^v{d}^h=-(d_p\otimes d_q),$$

while

$$d^h{d}^v=d_p\otimes d_q.$$

Therefore

$$d^v{d}^h=-d^h{d}^v.$$

So the anticommuting square in a double complex is exactly the Koszul interchange law applied to two odd differentials.

Equivalently,

$$d^v{d}^h+d^h{d}^v=0.$$

The two copies of $d^2=0$$d^2=0$ also give

$$(d^h{)}^2=0,$$

and

$$(d^v{)}^2=0.$$

Therefore

$${\mathsf{ChSk}}^{\sup }\hat{\otimes }{\mathsf{ChSk}}^{\sup }$$

is the signed sketch for anticommuting double complexes.

If $\mathcal{A}$$\mathcal A$ is an ordinary additive category, an anticommuting double complex in $\mathcal{A}$$\mathcal A$ is an additive functor

$$\mathrm{Und}({\mathsf{ChSk}}^{\sup }\hat{\otimes }{\mathsf{ChSk}}^{\sup })\to \mathcal{A}.$$

Thus the super structure is used to generate the Koszul sign, and then the resulting signed preadditive sketch may be interpreted in an ordinary additive category.

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$$\boxed{{\displaystyle \text{anticommuting double complex}=\text{model of the Koszul tensor square of the chain-complex sketch.}}}$$