\section{Tannakian Categories: definitions and motivation (notes by Christian Klevdal)}

Most of the material in this section may be found in Motives by Yves Andr\'{e}.

%Definition of rigid tensor categories
\begin{defn} A {\bf tensor category} over a commutative ring $F$ is an $F$-linear
category $\mc{C}$ that has a tensor structure, which consists of the following data:
\begin{enumerate}
\item A functor $\otimes \colon \mc{C} \times \mc{C} \to \mc{C}$. 
\item An identity object ${\bf 1}$.
\item Associativity and commutativity constraints, unit isomorphisms, which are
functorial isomorphisms 
\begin{align*}
a_{X,Y,Z} &\colon X \otimes (Y\otimes Z) \xrightarrow{\sim} (X\otimes Y)\otimes Z \\
c_{X,Y} &\colon X \otimes Y \xrightarrow{\sim} Y\otimes X \quad \text{such that }
c_{X,Y} = c_{Y,X}^{-1}\\
u_{X} &\colon X\otimes {\bf1} \xrightarrow{\sim} X, \quad u'_{X} \colon
{\bf1}\otimes X \xrightarrow{\sim} X 
\end{align*}
\end{enumerate}
These isomorphisms are required to satisfy compatibilities
\begin{itemize}
\item Associativity compatibility: The two isomorphisms
\[ X \otimes (Y \otimes (Z \otimes T)) \to ((X \otimes Y) \otimes Z) \otimes T \]
coming from applying associativity isomorphisms in various ways are equal. 
\item Compatibility of commutativity and associativity: The two isomorphisms
\[ X \otimes (Y \otimes Z) \to (Z \otimes X) \otimes Y \]
coming from applying associativity and commutativity isomorphisms in various ways
are equal. 
\item Compatibility of associativity and unit: The two isomorphisms
\[ X \otimes ({\bf1} \otimes Y) \to X \otimes Y \] 
coming from applying associativity and unit isomorphisms in various ways are equal. 
\end{itemize}
%Rigid tensor category
Further, the tensor category $(\mc{C}, \otimes)$ is said to be {\bf rigid} if:
\begin{enumerate}[resume]
\item There exists an autoduality ${}^\vee \colon \mc{C} \to \mc{C}^\text{op}$ such
that for each object $X$, the functor $\text{-}\otimes X^\vee$ is left adjoint to
$\text{-}\otimes X$ and $X^\vee \otimes \text{-}$ is right adjoint to $X \otimes
\text{-}$. 
\end{enumerate}
\end{defn}

%Definition of trace and rank
Note that if $(\mc{C}, \otimes)$ is a rigid tensor category then the unit and counit
of adjunction give maps $\varepsilon \colon X \otimes X^\vee \to {\bf 1}$ and $\eta
\colon {\bf 1} \to X^\vee \otimes X$ called evaluation and coevaluation
respectively. One can now define a {\bf trace map} from $\mathrm{End}(X)$ to the
commutative $F$-algebra $\mathrm{End}({\bf 1})$ sending $f \in \mathrm{End}(X)$ to
the composition
\[ {\bf 1} \xrightarrow{f} X^\vee \otimes X \xrightarrow{c_{X^\vee, X}} X\otimes
X^\vee \xrightarrow{\varepsilon} {\bf 1} \]
The first map is induced by $f$ using adjunction $\mathrm{Hom}(X,X) \cong
\mathrm{Hom}({\bf 1}, X^\vee \otimes X)$. We define the {\bf rank} of $X$ to be the
the trace of the identity, that is $\varepsilon \circ c_{X^\vee, X} \circ \eta \in
\mathrm{End}({\bf 1})$. 

%Examples of rigid tensor categories
\begin{exmpl}
If $F$ is a field and $G$ a group then the category of $\mathrm{Rep}_F(G)$ is a
rigid tensor category, where $\otimes$ is the usual tensor product,${\bf 1}$ is the
trivial representation and $V^\vee$ is the dual. Note $\mathrm{End}({\bf 1}) = F$
and the rank of an object $V$ is just the dimension of the representation.
\end{exmpl}

%Definition of tensor functor
\begin{defn}
A {\bf tensor functor} between tensor categories $(\mc{C}, \otimes_\mc{C}), (\mc{T},
\otimes_\mc{T})$ over $F$ is an $F$ linear functor $\omega \colon \mc{C} \to \mc{T}$
with functorial isomorphisms
\[ o_{X,Y} \colon \omega(X \otimes_\mc{C} Y) \xrightarrow{\sim} \omega(X)
\otimes_\mc{T} \omega(Y) \quad \quad o_1 \colon \omega({\bf 1}_\mc{C})
\xrightarrow{\sim} {\bf 1}_\mc{T} \]
that are compatible with constraints $a, c, u, u'$. 
\end{defn}

We can finally define what a (neutral) Tannakian category is. 

%Definition of Tannakian category
\begin{defn}
Suppose $(\mc{C}, \otimes)$ is a rigid abelian tensor category, i.e. $\mc{C}$  is
abelian and $\otimes$ is bi-additive. Let $K = \mathrm{End}({\bf 1})$. A {\bf fiber
functor} for $\mc{C}$ is an exact, faithful tensor functor 
\[ \omega \colon \mc{C} \to \mathrm{Vect}_L \]
to the category of finite dimensional $L$-vector spaces, where $L \supset K$. If
such a fiber functor exists, we say that $\mc{C}$ is {\bf Tannakian}. If $L = K$
then we say that $\mc{C}$ is {\bf neutral Tannakian}. 
\end{defn}

%Introduction to motives
\subsection{Homological Motives}
The interest in Tannakian categories in algebraic geometry comes from motives, which
we quickly review here. Much of the following is imprecise. For reference, one
should consult the book Motives by Yves Andr\'{e} or the article Classical Motives by
A.J. Scholl. For any field $k$, we can define the category of pure homological
motives over $k$ with coefficients in a field $F$. In order to define motives, we
first need to recall facts about algebraic cycles. Given a variety $X$ over $k$, and
a field $F$ denote $Z_F^r(X)$ to be the free $F$-module generated by irreducible
subvarieties of codimension $r$. An element of $Z_F^r(X)$ is called an algebraic
cycle. A Weil cohomology theory is a functor $H^\ast$ from smooth projective
varieties over $k$ to finite dimensional graded commutative $F$-algebras with a
cycle class map $Z_F^r(X) \to H^{2r}(X)$ satisfying Poincare duality and the Kunneth
formula (along with certain compatability and non degeneracy conditions). The
typical examples are $\ell$-adic cohomology, algebraic de Rham cohomology and if $k
\subset \mathbb{C}$ singular cohomology. If we fix a Weil cohomology theory, one can
define an equivalence relation called homological equivalence on $Z^r_F(X)$ by $Z
\sim_H W$ if they have the same image under the cycle class map. Define $A^r(X) =
Z^r_F(X)/\sim_H$. Given a map of varieties (which will mean smooth projective from
now on) $f \colon X \to Y$ there are pull back maps $f^\ast \colon A^r(Y) \to
A^r(X)$ and pushforward maps $f_\ast \colon A^r(X) \to A^{r + \mathrm{dim}(Y) -
\mathrm{dim}(X)}(Y)$. There is also a product structure $A^r(X)\otimes A^s(X) \to
A^{r + s}(X)$, denoted by $Z \otimes W \mapsto Z\cdot W$. For connected varieties
$X, Y$ we define the correspondences of degree $r$ to be $\mathrm{Corr}^r(X, Y) =
A^{\mathrm{dim}(X)+r}(X \times Y)$. We can define a composition law $\mathrm{Corr}^r(X,Y) \times
\mathrm{Corr}^s(Y,Z) \to \mathrm{Corr}^{r+s}(X,Z)$ given on cycles $Z \in
\mathrm{Corr}^r(X,Y), W \in \mathrm{Corr}^s(Y,Z)$ by 
\[ Z \circ W = (p_{XZ})_\ast(p_{XY}^\ast Z \cdot p_{YZ}^\ast W)\]
where the $p$ maps are the various projections from $X\times Y \times Z$. We may
finally define the category of pure homological motives $\mc{M}_k$ over $F$. The
objects are triples $(X, p, r)$ where $X$ is a (smooth projective) variety over $k$,
$p \in \mathrm{Corr}^0(X,X)$ is an idempotent correspondence and $r$ is an integer.
One should think of a motive as a `chunk of cohomology', corresponding to the image
 $pH^\ast(X)(r)$. For example,
if $X$ is smooth and projective and $x \in X$ one can check that $\{x\}\times X$
induces projection $H^\ast(X) \to H^0(X)$ for any Weil cohomology theory. In this
case we say that the $H^0$ projector is algebraic. It is conjectured that all (``K\:{u}nneth")
projectors $H^{\ast}(X) \to H^i(X)$ are algebraic.

We still need to define maps between motives. Given another motive $(Y, q, s)$, 
\[\mathrm{Hom}_{\mc{M}_k}((X,p,r), (Y,q,s)) = p\mathrm{Corr}^{s-r}(X,Y)q \]
Composition is given by composition of correspondences. There is a functor 
\[ h \colon \mathrm{Var}^{\mathrm{op}}_k \to \mc{M}_k, \quad \quad X \mapsto h(X) = (X, \Delta_X,
0) \] from the category of smooth projective varieties over $k$ to motives. A map
$\varphi \colon X \to Y$ is sent the the correspondence ${}^t\Gamma_\varphi$ the
transpose of the graph. We can define the sum of motives by 
\[(X,p,r) \oplus (Y, q, r) = (X \sqcup Y, p+q, r) \]
(we let the reader find the definition of sum for $r \neq s$) and the tensor product
\[(X,p, r) \otimes (Y, q, r) = (X \times Y, p\otimes q, r + s) \]
The identity motive is ${\bf 1} = (\mathrm{Spec} k, \mathrm{id}, 0)$. With the
obvious associativity, commutativity and unit isomorphisms, $\mc{M}_k$ is a tensor
category. If we define the dual of a motive $(X, p, r)$, where $X$ is purely $d$
dimensional, as 
\[ (X, p, r)^\vee = (X, {}^tp, d - r) \]
then $\mc{M}_k$ is a rigid tensor category. 

The hope is that $\mc{M}_k$ is a neutral Tannakian category. In order to show this,
two things must be true, first that $\mc{M}_k$ is abelian and second, that there is
a fiber functor. Focusing on the latter part, a natural choice of fiber functor
would be $H^\ast$ where $H^\ast$ is a Weil cohomology theory (above we stated that
Weil cohomology theories are functors from the category of smooth projective
varieties. The idea behind motives are to give a sort of `universal cohomology
theory', and indeed there is a unique way to extend $H^\ast$ to $\mc{M}_k$). Lets
see if $H^\ast$ is a tensor functor, which boils down to commutativity of the
following diagram. 
\begin{center}
        \begin{tikzcd}
        H^\ast(X)\otimes H^\ast(Y) \arrow[r, "\sim"] \arrow[d, "c_{\text{Vect}_F}"] &
H^\ast(X\times Y) \arrow[d, "c_{\mc{M}_k}"] \\
        H^\ast(Y)\otimes H^\ast(X) \arrow[r, "\sim"] & H^\ast(Y\times X)
        \end{tikzcd}
\end{center}
Tracing around the top, a tensor $\alpha \otimes \beta \in H^i(X) \otimes H^j(Y)$
gets sent to $p^\ast \alpha \cup q^\ast \beta$ where $X \xleftarrow{p} X\times Y
\xrightarrow{q} Y$ are the projections. Going around the bottom, $\alpha \otimes
\beta$ maps to $q^\ast \beta \otimes p^\ast \alpha$. Hence the diagram only commutes
up to the sign $(-1)^{ij}$! So in order for $H^\ast$ to be a fiber functor, one must
modify the commutativity constraint in $\mc{M}_k$ degree by degree and in order to
perform this, one needs to know that \emph{all} projectors $H^\ast(X) \to H^i(X)$
are algebraic. This is the Kunneth standard conjecture.

%Example: Motives over F_p
\begin{exmpl}
We will show (or at least provide evidence) that motives over finite fields $k$ (with the possible exception
of $\mathbb{F}_p \ldots$) with
coefficients in $\rats$ is not a \textit{neutral} Tannakian category, even though it is conjecturally
Tannakian. Again, this discussion
is imprecise and many details still should be checked. Consider $E$ a supersingular
elliptic curve such that $\mathrm{End}(E)$ is a quaternion algebra (here we mean endomorphisms
of the elliptic curve over $k$ itself--this is why $k=\mathbb{F}_p$ doesn't work in this example). We wish to
compute the rank of the motive $h^1(E) = (E, p, 0)$ where $p = \Delta_E -
\{0\}\times E - E \times \{0\}$ (this is the $H^1$ projector). For motives over
finite fields the Kunneth standard conjecture is known to be true, so we can modify
the commutativity constraint in such a way so that any Weil cohomology theory
$H^\ast$ is a tensor functor. The benefit of this is to aid in computing the rank of
the motive $h^1(E)$. Now one can check that a tensor functor between rigid tensor
categories will preserve rank so $\mathrm{rank}\, h^1(E) = \mathrm{rank}\,
H^\ast(h^1(E)) = \mathrm{rank}\, H^1(E)$ for any Weil cohomology theory. If we take
$H^\ast$ to be $\ell$-adic cohomology then $\mathrm{rank}\, H^1(E, \rats_\ell)$ is
the dimension of $H^1(E, \rats_\ell)$ which is just $2$. Hence $h^1(E)$ has rank 2.
If $\mc{M}_{k}$ is neutral Tannakian, then there exits a fiber functor 
\[ \omega \colon \mc{M}_{k} \to \mathrm{Vect}_\rats \]
From the above discussion, $\omega(h^1(E))$ is a 2 dimensional rational
representation. Since $\omega$ is faithful, $\mathrm{End}(E)$ embeds into
$\mathrm{End}(h^1(E))$. In other words, we have a faithful 2 dimensional rational
representation of a quaternion algebra. This is a contradiction since no such
representation can exist. 
\end{exmpl}

\section{The main theorem of neutral Tannakian categories (notes by Shiang Tang)}

Throughout this section, $G$ will be an affine group scheme over a field $k$. We denote by $\mathrm{Rep}(G)$ its category of finite-dimensional representations, and we let $\mathrm{Vec}_k$ denote the category of finite-dimensional $k$-vector spaces.
\textbf{Basic Question}: How to recognize a category with structure as $\mathrm{Rep}(G)$ for some affine group scheme $G$?

First, we ask the opposite question: Fix a field $k$, and fix an affine group scheme $G/k$, to what extent is $G$ determined by $\mathrm{Rep}(G)$?

\begin{exmpl}
Let $G$ be a finite group, regarded as a group scheme over $\mathbb{C}$. Then $\mathrm{Rep}(G)$ is an abelian category, better yet a semi-simple category. As an abelian category,
\[
\mathrm{Rep}(G) \cong \bigoplus_{\textrm{$W$ irreducible $G$-reps up to isomorphism}}\mathrm{Vec}_{\mathbb C}
\]
(Note that there are no nonzero maps between different summands.). This equivalence results by observing that
for all $V \in \mathrm{Rep}(G)$, 
\[
V=\bigoplus_{\textrm{irreducible $G$-reps $W$}}\left(\textrm{$W$-isotypic components of $V$}\right)
\]
(each component is canonically isomorphic to $\mathrm{Hom}_G(W,V)\otimes W$), and we have
\[
\mathrm{Hom}_G(W^{\otimes n},W^{\otimes m})\cong \mathrm{M}_{m,n}(\mathbb C).
\]
Thus we see that the underlying \textit{abelian} category $\mathrm{Rep}(G)$ does not remember $G$.
\end{exmpl}

\textbf{Key idea}: We can recover $G$ from the \emph{tensor abelian category} $\rep (G)$. 

\begin{ex}
\begin{enumerate}
\item Find two non-isomorphic finite groups that have "isomorphic character tables"? 
\item Interpret part (1) in light of the key idea.
\end{enumerate}
\end{ex}
\begin{proof}
First, we make it precise what it means by two groups having isomorphic character tables: Given a finite group $G$, denote by $\mathrm{Conj}(G)$ the set of conjugacy classes, and denote by $\mathrm{Char}(G)$ the set of irreducible characters. We say two finite groups $G, G'$ have isomorphic character table if there are bijections $\varphi: \mathrm{Conj}(G) \to \mathrm{Conj}(G')$, $\alpha:\mathrm{Char}(G) \to \mathrm{Char}(G')$ such that $\forall c\in \mathrm{Conj}(G), \forall \chi\in \mathrm{Char}(G)$,
$\chi(c)=\alpha(\chi)(\phi(c))$. For instance, it is easy to see that $D_4$ and $Q_8$ have isomorphic character tables: both of which have four one-dimensional character and a unique irreducible two-dimensional character. However, there difference will be seen immediately once we consider the alternating squares of their two-dimensional representations: one is isomorphic to the trivial character, the other is not. Therefore, $\mathrm{Rep}(D_4)$ is not isomorphic to $\mathrm{Rep}(Q_8)$ as abelian tensor categories, despite that $D_4$ and $Q_8$ have isomorphic character tables.
\end{proof}

\begin{defn}
Let $\aut^{\otimes}(\omega)$ be the functor from the category of $k$-algebras to the category of groups defined by:
\[
R \mapsto \aut^{\otimes}(\omega)(R),
\] 
where $\aut^{\otimes}(\omega)(R)$ consists of elements $\lambda=(\lambda_X)_{X\in\rep(G)}$ which is a collection of $R$-linear automorphisms $\lambda_X: \omega(X)\otimes_k R \to \omega(X)\otimes_k R$, for all $X \in \rep(G)$, satisfying
\begin{enumerate}
\item $\forall \phi:X\to Y$ in $\rep(G)$, the diagram
\begin{tikzcd}
\omega(X)\otimes_k R \arrow{r}{\phi} \arrow{d}{\lambda_X}
&\omega(Y)\otimes_k R \arrow{d}{\lambda_Y}\\
\omega(X)\otimes_k R \arrow{r}{\phi} 
&\omega(Y)\otimes_k R
\end{tikzcd} commutes.
\item For $X=\textrm{trivial rep}$, $\lambda_X=\mathrm{id}$.
\item $\lambda_{\omega(X_1)\otimes\omega(X_2)}=\lambda_{\omega(X_1)}\otimes\lambda_{\omega(X_2)}$, i.e., the following diagram commutes:

\begin{tikzcd}
(\omega(X_1)\otimes\omega(X_2))\otimes R \arrow{r}{\lambda_{\omega(X_1)\otimes\omega(X_2)}} \arrow{d}{\cong}
&(\omega(X_1)\otimes\omega(X_2))\otimes R \arrow{d}{\cong}\\
(\omega(X_1)\otimes R)\otimes (\omega(X_2)\otimes R) \arrow{r}{\lambda_{\omega(X_1)}\otimes\lambda_{\omega(X_2)}} 
&(\omega(X_1)\otimes R)\otimes (\omega(X_2)\otimes R)
\end{tikzcd}
\end{enumerate}

\end{defn}

\begin{thm}
Let $k$ be any field, and let $G/k$ be an affine group scheme. Let 
\[
\omega: \rep(G) \to \mathrm{Vec}_k
\] 
be the forgetful functor where $\rep(G)$ is the category of finite-dimensional representations of $G$ and $\mathrm{Vec}_k$ is the category of finite-dimensional vector spaces over $k$. Then the natural map 
\[
G \to \aut^{\otimes}(\omega)
\] 
of functors from the category of $k$-algebras to the category of groups is an isomorphism. Here the natural map is defined as follows:
for any $k$-algebra $R$, 

\[G(R) \to \aut^{\otimes}(\omega)(R)\]
\[g \mapsto \lambda_g=(X\otimes R \to X\otimes R)_{X\in \rep (G)}\]

where the linear automorphism $X\otimes R \to X\otimes R$ is induced by the action of $g$ on $X$.
\end{thm}

\begin{cor}
Let $G,G'$ be affine group schemes over $k$. Suppose 
\[F:\rep (G') \to \rep(G)\] is a tensor functor such that 
\begin{tikzcd}
\rep (G') \arrow{dr}{\omega_{G'}} \arrow{rr}{F}
&& \rep (G) \arrow{dl}{\omega_{G}} \\
& \mathrm{Vec}_k 
\end{tikzcd}
commutes. Then there exists a unique homomorphism $G \to G'$ inducing $F$.
\end{cor}
\begin{proof}
$F:\rep (G') \to \rep(G)$ induces a natural transformation $\Phi:\aut^{\otimes}(\omega_G) \to \aut^{\otimes}(\omega_{G'})$ given by

\[\Phi(R):\aut^{\otimes}(\omega_G)(R) \to \aut^{\otimes}(\omega_{G'})(R)\]
\[\lambda \mapsto (\lambda_{F(X)})_{X\in\rep(G')}\]
We need to check that the map is well-defined, i.e., $\lambda':=(\lambda_{F(X)})_{X\in\rep(G')}$ lands in $\aut^{\otimes}(\omega_{G'})(R)$. Item (1) of Definition 1.3 holds since $\omega_{G}(F(X))=\omega_{G'}(X)$ by assumption; item (2) obviously holds; since $F$ is a tensor functor, $\lambda'_{X}\otimes \lambda'_{Y}=\lambda_{F(X)}\otimes\lambda_{F(Y)}=\lambda_{F(X)\otimes F(Y)}=\lambda_{F(X\otimes Y)}=\lambda'_{X\otimes Y}$, Item (3) holds. $\Phi(R)$ is clearly a homomorphism. By Theorem 1.4, we have natural isomorphisms of functors
$\alpha:G \to \aut^{\otimes}(\omega_G)$, $\alpha':G' \to \aut^{\otimes}(\omega_G')$. Therefore, we obtain a homomorphism $\phi:=\alpha'^{-1}\circ \Phi \circ \alpha: G \to G'$.
\end{proof}

In a similar spirit, we can write down a dictionary between properties of $G$ and properties of $\rep (G)$. 

\begin{exmpl}
$G$ is algebraic and of finite type over $k$ is equivalent to $\rep (G)$ has a tensor generator, i.e., there exists an object $X$ in $\rep(G)$ such that every object of $\rep(G)$ is isomorphic to a subquotient of some 
\[\bigoplus_{\mathrm{m,n}} \left( X^{\otimes m} \otimes (X^{\vee})^{\otimes n} \right)\]

This may fail when $G$ is not algebraic, for instance, there is no such generator when $G=\mathrm{Gal}(\overline{\rats}/\rats)$, the absolute Galois group, which is a profinite group, and is not algebraic.
\end{exmpl}

Let us recall that our main task is going other way: having some category $\mc C$ that we want to recognize as $\rep(G)$.

\begin{thm}[Main Theorem]
Let $k$ be a field and let $(\mc C,\otimes)$ be a rigid abelian tensor category with $\mathrm{End}(\mathbf{1})=k$($\mathbf{1}$ is the unit object) and suppose that there exists a $k$-linear exact faithful tensor functor (called a ``fiber functor")
\[\omega: \mc C \to \mathrm{Vec}_k\]
Then 
\begin{enumerate}
\item The functor $\aut^{\otimes}(\omega)$ from the category of $k$-algebras to the category of groups is representable by an affine group scheme $G$ over $k$. 
\item $\omega$ induces an equivalence 
\[\mc C \xrightarrow{\sim} \rep(G)\] of tensor categories.
\end{enumerate}
\end{thm}
\begin{proof}
For a proof, see [Deligne-Milne, 2.11-2.16]. For part (2), $\omega$ induces a functor to $\mathrm{Rep}(G)$ since for all $X\in \mc C$, $\omega(X)$ acquires a $G$-action from the isomorphism $G\cong \aut^{\otimes}(\omega)$.
\end{proof}

