superalgebra and (synthetic ) supergeometry
A super vector space is an object in the non-trivial symmetric monoidal category structure on the monoidal category of $\mathbb{Z}/2$-graded vector spaces: as an object it is just a $\mathbb{Z}/2$-graded vector space, but the braiding of the underlying tensor product of vector spaces is taken to be the non-trivial linear map which on elements of homogeneous degree is given by
We make this precise as definition 8 below.
Super vector spaces form the basis of superalgebra (over ground rings which are fields) in direct analogy of how ordinary vector spaces form the basis of ordinary algebra. For more on this see below, and for yet more see at geometry of physics -- superalgebra.
For $k$ a field, we write $Vect_k$ for the category whose
objects are $k$-vector spaces;
morphisms are $k$-linear functions between these.
When the ground field $k$ is understood or when its precise nature is irrelevant, we will often notationally suppress it and speak of just the category Vect of vector spaces.
This is the category inside which linear algebra takes place.
Of course the category Vect has some special properties. Not only are its objects “linear spaces”, but the whole category inherits linear structure of sorts. This is traditionally captured by the following terminology for additive and abelian categories. Notice that there are several different but equivalent ways to state the following properties (discussed behind the relevant links).
Let $\mathcal{C}$ be a category.
Say that $\mathcal{C}$ has direct sums if it has finite products and finite coproducts and if the canonical comparison morphism between these is an isomorphism. We write $V \oplus W$ for the direct sum of two objects of $\mathcal{C}$.
Say that $\mathcal{C}$ is an additive category if it has direct sums and in addition it is enriched in abelian groups, meaning that every hom-set is equipped with the structure of an abelian group such that composition of morphisms is a bilinear map.
Say that $\mathcal{C}$ is an abelian category if it is an additive category and has property that its monomorphisms are precisely the inclusions of kernels and its epimorphisms are precisely the projections onto cokernels.
We also make the following definition of $k$-linear category, but notice that conventions differ as to which extra properties beyond Vect-enrichment to require on a linear category:
For $k$ a field (or more generally just a commutative ring), call a category $\mathcal{C}$ a $k$-linear category if
it is an abelian category (def. 2);
its hom-sets have the structure of $k$-vector spaces (generally $k$-modules) such that composition of morphisms in $\mathcal{C}$ is a bilinear map
and the underlying additive abelian group structure of these hom-spaces is that of the underlying abelian category.
In other words, a $k$-linear category is an abelian category with the additional structure of a Vect-enriched category (generally $k$Mod-enriched) such that the underlying Ab-enrichment according to def. 2 is obtained from the $Vect$-enrichment under the forgetful functor $Vect \to Ab$.
A functor between $k$-linear categories is called a $k$-linear functor if its component functions on hom-sets are linear maps with respect to the given $k$-linear structure, hence if it is a Vect-enriched functor.
The category Vect${}_k$ of vector spaces (def. 1) is a $k$-linear category according to def. 3.
Here the abstract direct sum is the usual direct sum of vector spaces, whence the name of the general concept.
For $V,W$ two $k$-vector spaces, the vector space structure on the hom-set $Hom_{Vect}(V,W)$ of linear maps $\phi \colon V \to W$ is given by “pointwise” multiplication and addition of functions:
for all $c_1, c_2 \in k$ and $\phi_1, \phi_2 \in Hom_{Vect}(V,W)$.
Recall the basic construction of the tensor product of vector spaces:
Given two vector spaces over some field $k$, $V_1, V_2 \in Vect_k$, their tensor product of vector spaces is the vector space denoted
whose elements are equivalence classes of tuples of elements $(v_1,v_2)$ with $v_i \in V_i$, for the equivalence relation given by
More abstractly this means that the tensor product of vector spaces is the vector space characterized by the fact that
it receives a bilinear map
(out of the Cartesian product of the underlying sets)
any other bilinear map of the form
factors through the above bilinear map via a unique linear map
The existence of the tensor product of vector spaces, def. 4, equips the category Vect of vector spaces with extra structure, which is a “categorification” of the familiar structure of a semi-group. One also says “monoid” for semi-group and therefore categories equipped with a tensor product operation are also called monoidal categories:
A monoidal category is a category $\mathcal{C}$ equipped with
a functor
out of the product category of $\mathcal{C}$ with itself, called the tensor product,
an object
called the unit object or tensor unit,
called the associator,
called the left unitor, and a natural isomorphism
called the right unitor,
such that the following two kinds of diagrams commute, for all objects involved:
triangle identity:
the pentagon identity:
As expected, we have the following basic example:
For $k$ a field, the category Vect${}_k$ of $k$-vector spaces becomes a monoidal category (def. 5) as follows
the abstract tensor product is the tensor product of vector spaces $\otimes_k$ from def. 4;
the tensor unit is the field $k$ itself, regarded as a 1-dimensional vector space over itself;
the associator is the map that on representing tuples acts as
the left unitor is the map that on representing tuples is given by
and the right unitor is similarly given by
That this satisifes the pentagon identity (def. 5) and the left and right unit identities is immediate on representing tuples.
But the point of the abstract definition of monoidal categories is that there are also more exotic examples. The followig one is just a minimal enrichment of example 2, and yet it will be important.
Let $G$ be a group (or in fact just a monoid/semi-group). A $G$-graded vector space $V$ is a direct sum of vector spaces labeled by the elements in $G$:
of $G$-graded vector spaces is a linear map that respects this direct sum structure, hence equivalently a direct sum of linear maps
for all $g \in G$, such that
This defines a category, denoted $Vect^G$. Equip this category with a tensor product which on the underlying vector spaces is just the tensor product of vector spaces from def. 4, equipped with the $G$-grading which is obtained by multiplying degree labels in $G$:
The tensor unit for the tensor product is the ground field $k$, regarded as being in the degree of the neutral element $e \in G$
The associator and unitors are just those of the monoidal structure on plain vector spaces, from example 2.
One advantage of abstracting the concept of a monoidal category is that it allows to prove general statements uniformly for all kinds of tensor products, familar ones and more exotic ones. The following lemma 1 and remark 1 are two important such statements.
(Kelly 64)
Let $(\mathcal{C}, \otimes, 1)$ be a monoidal category, def. 5. Then the left and right unitors $\ell$ and $r$ satisfy the following conditions:
$\ell_1 = r_1 \;\colon\; 1 \otimes 1 \overset{\simeq}{\longrightarrow} 1$;
for all objects $x,y \in \mathcal{C}$ the following diagrams commutes:
and
For proof see at monoidal category this lemma and this lemma.
Just as for an associative algebra it is sufficient to demand $1 a = a$ and $a 1 = a$ and $(a b) c = a (b c)$ in order to have that expressions of arbitrary length may be re-bracketed at will, so there is a coherence theorem for monoidal categories which states that all ways of freely composing the unitors and associators in a monoidal category (def. 5) to go from one expression to another will coincide. Accordingly, much as one may drop the notation for the bracketing in an associative algebra altogether, so one may, with due care, reason about monoidal categories without always making all unitors and associators explicit.
(Here the qualifier “freely” means informally that we must not use any non-formal identification between objects, and formally it means that the diagram in question must be in the image of a strong monoidal functor from a free monoidal category. For example if in a particular monoidal category it so happens that the object $X \otimes (Y \otimes Z)$ is actually equal to $(X \otimes Y)\otimes Z$, then the various ways of going from one expression to another using only associators and this “accidental” equality no longer need to coincide.)
The above discussion makes it clear that a monoidal category is like a monoid/semi-group, but “categorified”. Accordingly we may consider additional properties of monoids/semi-groups and correspondingly lift them to monoidal categories. A key such property is commutativity. But while for a monoid commutativity is just an extra property, for a monoidal category it involves choices of commutativity-isomorphisms and hence is extra structure. We will see below that this is the very source of superalgebra.
The categorification of “commutativity” comes in two stages: braiding and symmetric braiding.
A braided monoidal category, is a monoidal category $\mathcal{C}$ (def. 5) equipped with a natural isomorphism
(for all objects $x,y in \mathcal{C}$) called the braiding, such that the following two kinds of diagrams commute for all objects involved (“hexagon identities”):
and
where $a_{x,y,z} \colon (x \otimes y) \otimes z \to x \otimes (y \otimes z)$ denotes the components of the associator of $\mathcal{C}^\otimes$.
A symmetric monoidal category is a braided monoidal category (def. 6) for which the braiding
satisfies the condition:
for all objects $x, y$
In analogy to the coherence theorem for monoidal categories (remark 1) there is a coherence theorem for symmetric monoidal categories (def. 7), saying that every diagram built freely (see remark 2) from associators, unitors and braidings such that both sides of the diagram correspond to the same permutation of objects, coincide.
Consider the simplest non-trivial special case of $G$-graded vector spaces from example 3, the case where $G = \mathbb{Z}/2$ is the cyclic group of order two.
A $\mathbb{Z}/2$-graded vector space is a direct sum of two vector spaces
where we think of $V_{even}$ as the summand that is graded by the neutral element in $\mathbb{Z}/2$, and of $V_{odd}$ as being the summand that is graded by the single non-trivial element.
A homomorphism of $\mathbb{Z}/2$-graded vector spaces
is a linear map of the underlying vector spaces that respects the grading, hence equivalently a pair of linear maps
between then summands in even degree and in odd degree, respectively:
The tensor product of $\mathbb{Z}/2$-graded vector space is the tensor product of vector spaces of the underlying vector spaces, but with the grading obtained from multiplying the original gradings in $\mathbb{Z}/2$. Hence
and
As in example 3, this definition makes $\mathbb{Z}/2$ a monoidal category def. 5.
There are, up to braided monoidal equivalence of categories, precisely two choices for a symmetric braiding (def. 7)
on the monoidal category $(Vect_k^{\mathbb{Z}/2}, \otimes_k)$ of $\mathbb{Z}/2$-graded vector spaces from def. 4:
the trivial braiding which is the natural linear map given on tuples $(v_1,v_2)$ representing an element in $V_1 \otimes V_2$ (according to def. 4) by
the super-braiding which is the natural linear function given on tuples $(v_1,v_2)$ of homogeneous degree (i.e. $v_i \in (V_i)_{\sigma_i} \hookrightarrow V_i$, for $\sigma_i \in \mathbb{Z}/2$) by
For $(\mathcal{C}, \otimes, 1)$ a monoidal category, write
for the full subcategory on those $L \in \mathcal{C}$ which are invertible objects under the tensor product, i.e. such that there is an object $L^{-1} \in \mathcal{C}$ with $L \otimes L^{-1} \simeq 1$ and $L^{-1} \otimes L \simeq 1$. Since the tensor unit is clearly in $Line(L)$ (with $1^{-1} \simeq 1$) and since with $L_1, L_2 \in Line(\mathcal{C}) \hookrightarrow \mathcal{C}$ also $L_1 \otimes L_2 \in Line(\mathcal{C})$ (with $(L_1 \otimes L_2)^{-1} \simeq L_2^{-1} \otimes L_1^{-1}$) the monoidal category structure on $\mathcal{C}$ restricts to $Line(\mathcal{C})$.
Accordingly any braiding on $(\mathcal{C}, \otimes,1)$ restricts to a braiding on $(Line(\mathcal{C}), \otimes, 1)$. Hence it is sufficient to show that there is an essentially unique non-trivial symmetric braiding on $(Line(\mathcal{C}), \otimes, 1)$, and that this is the restriction of a braiding on $(\mathcal{C}, \otimes, 1)$.
Now $(Line(\mathcal{C}, \otimes , 1))$ is necessarily a groupoid (the “Picard groupoid” of $\mathcal{C}$) and in fact is what is called a 2-group. As such we may regard it equivalently as a homotopy 1-type with group structure, and as such it it is equivalent to its delooping
regarded as a pointed homotopy type. (See at looping and delooping).
The Grothendieck group of $(\mathcal{C}, \otimes, 1)$ is
the fundamental group of the delooping space.
Now a symmetric braiding on $Line(\mathcal{C})$ is precisely the structure that makes it a symmetric 2-group which is equivalently the structure of a second delooping $B^2 Line(\mathcal{C})$ (for the braiding) and then a third delooping $B^3 Line(\mathcal{C})$ (for the symmetry), regarded as a pointed homotopy type.
This way we have rephrased the question equivalently as a question about the possible k-invariants of spaces of this form.
Now in the case at hand, $Line(Vect^{\mathbb{Z}/2})$ has precisely two isomorphism classes of objects, namely the ground field $k$ itself, regarded as being in even degree and regarded as being in odd degree. We write $k^{1\vert 0}$ and $k^{0 \vert 1}$ for these, respectively. By the rules of the tensor product of graded vector spaces we have
and
In other words
Now under the above homotopical identification the non-trivial braiding is identified with the elements
Due to the symmetry condition (def. 7) we have
which implies that
Therefore for classifying just the symmetric braidings, it is sufficient to restrict the hom-spaces in $Line(Vect^{\mathbb{Z}/2})$ from being either $k$ or empty, to hom-sets being $\mathbb{Z}/2 = \{+1-1\} \hookrightarrow k$ or empty. Write $\widetilde{Line}(sVect)$ for the resulting 2-group.
In conclusion then the equivalence classes of possible k-invariants of $B^3 Line(sVect)$, hence the possible symmetric braiding on $Line(Vect^{\mathbb{Z}/2})$ are in the degree-4 ordinary cohomology of the Eilenberg-MacLane space $K(\mathbb{Z}/2,3)$ with coefficients in $\mathbb{Z}/2$. One finds (…)
The symmetric monoidal category (def. 7)
whose underlying monoidal category is that of $\mathbb{Z}/2$-graded vector spaces (example 4);
whose braiding (def. 6) is the unique non-trivial symmtric grading $\tau^{super}$ from prop. 1 is called the category of super vector spaces
The non-full symmetric monoidal subcategory
of
(on the two objects $k^{1\vert 0}$ and $k^{0\vert 1}$ and with hom-sets restricted to $\{+1,-1\} \subset k$, as in the proof of prop. 1) happens to be the 1-truncation of the looping of the sphere spectrum $\mathbb{S}$, regarded as a group-like E-infinity space (“abelian infinity-group”)
It has been suggested (in Kapranov 15) that this and other phenomena are evidence that in the wider context of homotopy theory/stable homotopy theory super-grading (and hence superalgebra) is to be regarded as but a shadow of grading in higher algebra over the sphere spectrum. Notice that the sphere spectrum is just the analog of the group of integers in stable homotopy theory.
By internalizing algebra and geometry in the category $sVect$ of super vector spaces, one obtains the corresponding superalgebra and supergeometry.
For example
monoids in $sVect$ are super algebras and, more interestingly, commutative monoids in $sVect$ are supercommutative superalgebra;
hence for example commutative Hopf algebras in $sVect$ are equivalently supercommutative Hopf algebras, which are the formal duals of affine algebraic supergroups;
Lie algebras in $sVect$ are super Lie algebras;
manifolds modeled on objects of $sVect$ are supermanifolds
etc.
By the above definition, any structure in $sVect$ works just like the corresponding structure in Vect, but with a sign inserted whenever two odd-graded symbols are interchanged. For more on this see also at signs in supergeometry.
Deligne's theorem on tensor categories says that all suitable tensor categories of subexponential growth have a fiber functor to $sVect$ and are equivalent to categories of representations of affine algebraic supergroups.
Veeravalli Varadarajan, section 3.1 of Supersymmetry for mathematicians: An introduction, Courant lecture notes in mathematics, American Mathematical Society Providence, R.I 2004
Dennis Westra, section 3 of Superrings and supergroups, 2009 (pdf)