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\theoremstyle{plain}
\newtheorem{dualspacethm}{Theorem}
\newtheorem{dualbasisthm}[dualspacethm]{Theorem}
\newtheorem{vsisothm}[dualspacethm]{Theorem}
\newtheorem{linearmapthm}[dualspacethm]{Theorem}
\newtheorem{dualtensorthm}[dualspacethm]{Theorem}
\newtheorem{dualarrowthm}[dualspacethm]{Theorem}
\newtheorem{frobeniusform}[dualspacethm]{Theorem}

\theoremstyle{definition}
\newtheorem{algebradef}{Definition}
\newtheorem{coalgebradef}[algebradef]{Definition}
\newtheorem{frobeniusdef}[algebradef]{Definition}
\newtheorem{dualspacedef}[algebradef]{Definition}
\newtheorem{dualarrowdef}[algebradef]{Definition}
\newtheorem{complexnum}{Example}
\newtheorem{groupalgebra}[complexnum]{Example}
\newtheorem{squarematrix}[complexnum]{Example}

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\title{Constructing Frobenius Algebras}

\author{Dylan G.L. Allegretti}

\date{August 17, 2007}

\begin{document}

\begin{abstract}
We discuss the relationship between algebras, coalgebras, and Frobenius algebras. We describe a method of constructing Frobenius algebras, given certain finite-dimensional algebras, and we demonstrate the method with several concrete examples.
\end{abstract}

\maketitle


\section{Introduction}

This paper was written for the 2007 summer math REU at the University of Chicago. It describes algebraic structures called Frobenius algebras and explains some of their basic properties. To make the paper accessible to as many readers as possible, we have included definitions of all the most important concepts. We only assume that the reader is familiar with linear algebra and basic terminology from category theory. There are many introductory texts on linear algebra; for an introduction to category theory, see the first chapter of \cite{Mac Lane 1988}.

Our discussion begins with the definition of an algebra, a coalgebra, and a Frobenius algebra. Then we show how to explicitly construct a coalgebra given a finite-dimensional algebra. We show that the resulting structure is closely related to a Frobenius algebra. The paper concludes with several examples.



\section{Algebras, Coalgebras, and Frobenius Algebras}

In this section we define the three main algebraic structures we wish to consider.

\begin{algebradef}
\label{def:algebradef}
An \textit{algebra} $A$ over a field $K$ is a vector space over $K$ together with a $K$-linear vector multiplication $\mu:A\otimes A\rightarrow A$, $x\otimes y\mapsto xy$ and a $K$-linear unit map $\eta:K\rightarrow A$ such that the following three diagrams commute.
\begin{gather*}
\xymatrix{
A\otimes A\otimes A \ar[r]^-{\textnormal{id}_A\otimes\mu} \ar[d]_-{\mu\otimes\textnormal{id}_A} & A\otimes A \ar[d]^-{\mu}\\
A\otimes A \ar[r]_-{\mu} & A
} \\
\xymatrix{
K\otimes A \ar[r]^-{\eta\otimes\textnormal{id}_A} \ar[rd] & A\otimes A \ar[d]^-{\mu}\\
& A
}
\qquad \xymatrix{
A\otimes A \ar[d]_-{\mu} & A\otimes K \ar[l]_-{\textnormal{id}_A\otimes\eta} \ar[ld]\\
A
}
\end{gather*}
The rectangular diagram expresses the fact that multiplication is associative, and the two triangular diagrams express the \textit{unit condition}.
\end{algebradef}



If we reverse all arrows in the above diagrams, we obtain the axioms of a coalgebra.



\begin{coalgebradef}
\label{def:coalgebradef}
A \textit{coalgebra} $A$ over a field $K$ is a vector space over $K$ together with two $K$-linear maps $\delta:A\rightarrow A\otimes A$ and $\varepsilon:A\rightarrow K$ such that the following three diagrams commute.
\begin{gather*}
\xymatrix{
A\otimes A\otimes A & \ar[l]_-{\textnormal{id}_A\otimes\delta} A\otimes A\\
A\otimes A \ar[u]^-{\delta\otimes\textnormal{id}_A} & \ar[l]^-{\delta} \ar[u]_-{\delta} A
} \\
\xymatrix{
K\otimes A & \ar[l]_-{\varepsilon\otimes\textnormal{id}_A} A\otimes A \\
& \ar[lu] A \ar[u]_-{\delta}
}
\qquad \xymatrix{
A\otimes A \ar[r]^-{\textnormal{id}_A\otimes\varepsilon} & A\otimes K \\
A \ar[u]^-{\delta} \ar[ru]
}
\end{gather*}
The map $\delta$ is called \textit{comultiplication}, and the map $\varepsilon$ is called the \textit{counit}. The rectangular diagram expresses a property called \textit{coassociativity}, and the two triangular diagrams express the \textit{counit condition}.
\end{coalgebradef}



\begin{frobeniusdef}
A \textit{Frobenius algebra} is a finite-dimensional algebra $A$ over a field $K$ together with a map $\sigma :  A \times A \rightarrow K$ which satisfies
\begin{align*}
& \sigma (x y, z) = \sigma (x, y z) \\
& \sigma (x_1 + x_2 , y) = \sigma (x_1 , y) + \sigma (x_2 , y) \\
& \sigma (x , y_1 + y_2) = \sigma (x , y_1) + \sigma (x , y_2) \\
& \sigma (a x , y) = \sigma (x , a y) = a \sigma (x , y)
\end{align*}
for all $x$, $y$, $z$, $x_1$, $x_2$, $y_1$, $y_2\in A$, \ $a\in K$, and $\sigma (x,y)=0$ for all $x$ only if  $y=0$. This last condition is called \textit{nondegeneracy}, and the map $\sigma$ is called the \textit{Frobenius form} of the algebra.
\end{frobeniusdef}



\section{Dual Spaces and Dual Maps}

In this section we show how to explicitly construct a coalgebra given a finite-dimensional algebra. If we then compose the multiplication map with the counit map, we obtain a map which satisfies all of the axioms of a Frobenius form, except possibly nondegeneracy.



\begin{dualspacedef}
If $V$ is a vector space over a field $K$, then the \textit{dual space} denoted $V^*$ is the set of $K$-linear maps $V\rightarrow K$.
\end{dualspacedef}



\begin{dualspacethm}
If $V$ is a finite-dimensional vector space over a field $K$, then $V^*$ is a vector space over $K$.
\end{dualspacethm}

\begin{proof}
Clearly $V^*$ is closed with respect to addition. Since $K$ is a field, the addition operation is associative and commutative. The zero element of $K$ is an identity for $V^*$, and the inverse of a vector $v^*\in V^*$ is simply $\textbf{-1} v^*$ where $\textbf{-1}$ is the additive inverse of the unit element in $K$. Hence $V^*$ is an abelian group with respect to addition.

Let $v^*$, $v_1^*$, and~$v_2^*$ be maps in $V^*$. Since these maps are $K$-linear and $K$ is a field, we have
\begin{align*}
& av^*(x)=v^*(ax)\in V \\
& a(v_1^*+v_2^*) = av_1^*+av_2^* \\
& (a+b)v^*(x) = v^*((a+b)x) = v^*(ax+bx) \\
& \qquad = v^*(ax)+v^*(bx) = av^*(x)+bv^*(x) \\
& a(bv^*(x)) = (ab)v^*(x)
\end{align*}
for all $x\in V$, \ $a$, $b\in K$, so $V^*$ satisfies all of the axioms of a vector space.
\end{proof}



\begin{dualbasisthm}
\label{thm:dualbasisthm}
If $V$ has a basis {$\textbf{e}_1,\dots,\textbf{e}_n$}, then the map $\textbf{e}_i^*$ defined by
\[
\label{basisdef}
	\textbf{e}_i^* (\sum_{j=1}^n c_j \textbf{e}_j) = c_i
\tag{$*$}
\]
is linear. Moreover, the $\textbf{e}_i^*$ form a basis for $V^*$.
\end{dualbasisthm}

\begin{proof}
We have
\begin{align*}
& \textbf{e}_i^* (\sum_{j=1}^n c_j \textbf{e}_j + \sum_{j=1}^n d_j \textbf{e}_j) = \textbf{e}_i^* (\sum_{j=1}^n (c_j + d_j) \textbf{e}_j)	\\
& \qquad = c_i + d_i = \textbf{e}_i^* (\sum_{j=1}^n c_j \textbf{e}_j) + \textbf{e}_i^* (\sum_{j=1}^n d_j \textbf{e}_j)
\end{align*}
and
\[
\textbf{e}_i^* (a \sum_{j=1}^n c_j \textbf{e}_j) = \textbf{e}_i^* (\sum_{j=1}^n a c_j \textbf{e}_j) = a c_i = a \textbf{e}_i^* (\sum_{j=1}^n c_j \textbf{e}_j),
\]
so the map is linear. Now if $\alpha_1\textbf{e}_1^*(x)+\dots+\alpha_n\textbf{e}_n^*(x)=0$ for all $x$, then surely $\alpha_1,\dots,\alpha_n=0$, for if $x=\textbf{e}_i$ then $\alpha_1\textbf{e}_1^*(x)+\dots+\alpha_n\textbf{e}_n^*(x)=\alpha_i$. This shows that the $\textbf{e}_i^*$ are linearly independent. The set {$\textbf{e}_1^*,\dots,\textbf{e}_n^*$} is clearly a spanning set for $V^*$, so it is a basis for $V^*$.
\end{proof}



\begin{vsisothm}
\label{thm:vsisothm}
The map
\begin{align*}
\varphi_V:V & \rightarrow V^* \\
\textbf{e}_i & \mapsto e_i^*
\end{align*}
is a vector space isomorphism.
\end{vsisothm}

\begin{proof}
Equation \thetag{$*$} implies a one-one correspondence between the basis vectors of $V$ and the basis maps of $V^*$. Since each map in $V^*$ can be represented as a \textit{unique} linear combination of the basis maps $\textbf{e}_1^*,\dots,\textbf{e}_n^*$, this implies that $\varphi_V$ is injective. Since the basis maps $\textbf{e}_1^*,\dots,\textbf{e}_n^*$ span $V^*$, the map $\varphi_V$ is also surjective. By Theorem \ref{thm:dualbasisthm}, the map $\varphi_V$ is linear, so $\varphi_V$ is a vector space isomorphism.
\end{proof}



\begin{dualtensorthm}
\label{thm:dualtensorthm}
If $V$ and $W$ are finite-dimensional vector spaces, then there is an isomorphism ${(V\otimes W)}^*\cong V^*\otimes W^*$.
\end{dualtensorthm}

\begin{proof}
Since $V$ and $W$ are finite-dimensional, Theorem \ref{thm:vsisothm} says that the maps $\varphi_V:V\rightarrow V^*$ and $\varphi_W:W\rightarrow W^*$ are isomorphisms. It follows that the tensor product
	\[\varphi_V\otimes \varphi_W:V\otimes W \rightarrow V^*\otimes W^*
\]
is an isomorphism. Theorem \ref{thm:vsisothm} also implies that ${(V\otimes W)}^*\cong V\otimes W$. Thus
	\[{(V\otimes W)}^*\cong V\otimes W\stackrel{\cong}{\rightarrow}V^*\otimes W^*,
\]
and since the isomorphism is transitive, we have ${(V\otimes W)}^*\cong V^*\otimes W^*$.
\end{proof}



\begin{dualarrowdef}
Suppose that $V$ and $W$ are vector spaces over $K$. If $\alpha:V\rightarrow W$ is a $K$-linear map, then we write
\begin{align*}
\alpha^*:W^* & \rightarrow V^* \\
f & \mapsto f\circ\alpha.
\end{align*}
and call $\alpha^*$ the \textit{dual map}. Note that $\alpha$ sends $V$ into $W$ while $\alpha^*$ sends $W^*$ into $V^*$. In this sense the dual map reverses arrows.
\end{dualarrowdef}



\begin{dualarrowthm}
\label{thm:dualarrowthm}
Suppose that $V$, $W$, and~$X$ are vector spaces over $K$. If $\alpha:V\rightarrow W$ is a $K$-linear map, then $\alpha^*$ is a $K$-linear map. Moreover, if $a:V\rightarrow W$ and $b:W\rightarrow X$ are $K$-linear maps, then ${(b\circ a)}^* = a^*\circ b^*$.
\end{dualarrowthm}

\begin{proof}
If $\alpha$ is a $K$-linear map, then
\[
\alpha^* (f_1 + f_2) = (f_1+f_2)(\alpha) = f_1(\alpha) + f_2(\alpha) = \alpha^*(f_1) + \alpha^*(f_2)
\]
and
	\[\alpha^*(af) = af(\alpha) = a\alpha^*(f)
\]
for all $f$, $f_1$, $f_2\in W^*$, \ $a\in K$, so $\alpha^*$ is a $K$-linear map. For any map $f:X\rightarrow K$, we also have ${(b\circ a)}^*\circ f = f\circ (b\circ a) = (f\circ b)\circ a = a^*\circ(f\circ b) = a^*\circ(b^*\circ f)$.
\end{proof}



We are now in a position to construct a coalgebra from an algebra. Let $A$ be a finite-dimensional algebra over a field $K$. If we replace each vector space in Definition \ref{def:algebradef} with its corresponding dual space, and if we replace the maps with their corresponding dual maps, then we obtain the following three commutative diagrams.
\begin{gather*}
\xymatrix{
{(A\otimes A\otimes A)}^* & \ar[l]_-{{(\textnormal{id}_A\otimes\mu)}^*} {(A\otimes A)}^*\\
{(A\otimes A)}^* \ar[u]^-{{(\mu\otimes\textnormal{id}_A)}^*} & \ar[l]^-{\mu^*} \ar[u]_-{\mu^*} A^*
} \\
\xymatrix{
{(K\otimes A)}^* & \ar[l]_-{{(\eta\otimes\textnormal{id}_A)}^*} {(A\otimes A)}^* \\
& \ar[lu] A^* \ar[u]_-{\mu^*}
}
\qquad \xymatrix{
{(A\otimes A)}^* \ar[r]^-{{(\textnormal{id}_A\otimes\eta)}^*} & {(A\otimes K)}^* \\
A^* \ar[u]^-{\mu^*} \ar[ru]
}
\end{gather*}
If we apply Theorems \ref{thm:dualtensorthm} and \ref{thm:dualarrowthm}, we obtain a coalgebra $A^*$ with comultiplication $\mu^*$ and counit $\eta^*$. Fix an isomorphism $\varphi:A\stackrel{\cong}{\rightarrow} A^*$ on basis elements by $\varphi(e_i) = e_i^*$, where the first basis element is the unit element $\textbf{1} = \eta(1)$. This map transforms the coalgebra $A^*$ into a coalgebra $A$ with counit $\varepsilon = \eta^*\circ \varphi: A \rightarrow K$.



The following theorem shows how this construction relates to Frobenius algebras.



\begin{frobeniusform}
\label{thm:frobeniusform}
The map $\varepsilon\circ\mu$ has all of the properties of a Frobenius form, except possibly nondegeneracy.
\end{frobeniusform}

\begin{proof}
By the definition of $\varepsilon$ and $\mu$ we have
	\[(\varepsilon\circ\mu)(x y,z)=\varepsilon(x y z)=(\varepsilon\circ\mu)(x,y z)
\]
for all $x$, $y$, $z\in A$. Since $\varepsilon$ is linear, we also have
\begin{align*}
(\varepsilon\circ\mu)(x_1+x_2,y) & =\varepsilon((x_1+x_2) y)=\varepsilon(x_1 y+x_2 y)\\
& =\varepsilon(x_1 y)+\varepsilon(x_2 y)=(\varepsilon\circ\mu)(x_1,y)+(\varepsilon\circ\mu)(x_2,y)
\end{align*}
and
\[
(\varepsilon\circ\mu)(ax,y)=\varepsilon(axy)=a\varepsilon(x y)=a(\varepsilon\circ\mu)(x,y),
\]
and similarly, $(\varepsilon\circ\mu)(x,y_1+y_2)=(\varepsilon\circ\mu)(x,y_1)+(\varepsilon\circ\mu)(x,y_2)$ and $(\varepsilon\circ\mu)(x,ay)=a(\varepsilon\circ\mu)(x,y)$ for all $x$, $y$, $z$, $x_1$, $x_2$, $y_1$, $y_2\in A$, \ $a\in K$. This shows that $\varepsilon\circ\mu$ has all of the properties of a Frobenius form, except possibly nondegeneracy.
\end{proof}



Conversely, one can show that every Frobenius algebra has both algebra and coalgebra structure. This property of Frobenius algebras allows us to define ``topological quantum field theories'', which are useful in topology and physics. For more information on topological quantum field theories, see \cite{Kock 2003}.



\section{Examples}

\begin{complexnum}
One can easily check that the field of complex numbers together with the inclusion map $\eta:\mathbb{R}\hookrightarrow\mathbb{C}$ and ordinary multiplication $\mu:\mathbb{C}\otimes\mathbb{C}\rightarrow\mathbb{C}$ is a finite-dimensional algebra. Choose the canonical basis $\left\{\textbf{e}_1=1,\textbf{e}_2=i\right\}$ for $\mathbb{C}$. Then equation~\thetag{$*$} determines the basis maps for $\mathbb{C}^*$. Since the unit map $\eta$ is simply the inclusion, we have
\begin{align*}
& \eta^*(\textbf{e}_1^*)=\textbf{e}_1^*(\eta)=\textnormal{id}_\mathbb{R} \\
& \eta^*(\textbf{e}_2^*)=\textbf{e}_2^*(\eta)=0.
\end{align*}
This proves that $\varepsilon=\Re$ where $\Re$ is the ``real part function'' defined by $\Re(x+iy)=x$ for real $x$ and $y$. The map $\sigma(x,y)=\Re(x y)$ is nondegenerate, so it is a Frobenius form for the algebra.
\end{complexnum}



\begin{groupalgebra}
Suppose that $G=\{t_0,\dots,t_n\}$ is a finite group written multiplicatively with identity element $t_0$. One can show that the set $K[G]$ of formal linear combinations $\sum c_it_i$ \ ($c_i\in K$) together with the unit map
\begin{align*}
\eta:K & \rightarrow K[G] \\
a & \mapsto at_0,
\end{align*}
and multiplication $\mu$ given by multiplication in $G$, is a finite-dimensional algebra. Since $\{t_0,\dots,t_n\}$ is a basis for $K[G]$, equation~\thetag{$*$} gives a basis $\{t_0^*,\dots,t_n^*\}$ for ${K[G]}^*$. Then we have
\begin{align*}
& \eta^*(t_0^*)=t_0^*(\eta)=\textnormal{id}_K \\
& \eta^*(t_0^*)=t_i^*(\eta)=0 \ (i\neq0).
\end{align*}
By the definition of $\varepsilon$, we have
\begin{align*}
\varepsilon: K[G]&\rightarrow K \\
t_0&\mapsto 1 \\
t_i&\mapsto 0 \ (i\neq0).
\end{align*}
Finally, since the composite map $\sigma=\varepsilon\circ\mu$ is nondegenerate, it is a Frobenius form for the algebra.
\end{groupalgebra}



\begin{squarematrix}
It is again easy to check that the set $\textnormal{Mat}_n (K)$ of $n\times n$ matrices over a field $K$, together with the unit map
\begin{align*}
\eta:K & \rightarrow \textnormal{Mat}_n (K) \\
a & \mapsto
\begin{pmatrix}
	a & 0 & \dots & 0 \\
	0 & a & \dots & 0 \\
	\vdots & \vdots & \ddots & \vdots \\
	0 & 0 & \dots & a
\end{pmatrix}
\end{align*}
and ordinary matrix multiplication $\mu:\textnormal{Mat}_n (K)\otimes \textnormal{Mat}_n (K)\rightarrow \textnormal{Mat}_n (K)$, is a finite-dimensional algebra. Choose the canonical basis $\{\textbf{e}_{1,1},\dots,\textbf{e}_{n,n}\}$ for ${\textnormal{Mat}_n (K)}$ in which each matrix $\textbf{e}_{i,j}$ contains a $1$ in the $(i,j)$ position and $0$'s everywhere else. Then equation~\thetag{$*$} determines a basis $\{\textbf{e}_{1,1}^*,\dots,\textbf{e}_{n,n}^*\}$ for ${\textnormal{Mat}_n (K)}^*$. Since $\eta^*(\textbf{e}_{ij}^*)=\textbf{e}_{ij}^*(\eta)$ and $\eta(1)=\sum_i \textbf{e}_{ii}$, it follows that
\[
\eta^*(\textbf{e}_{ij}^*)=
\begin{cases}
1, & \text{$i=j$} \\
0 & \text{$i\neq j$.}
\end{cases}
\]
By forming linear combinations of these terms, we see that the counit is the \textit{matrix trace} obtained by summing all of the coordinates along the main diagonal of a matrix:
	\[\varepsilon(X) = \textnormal{tr} (X) = \sum_i (X)_{ii}.
\]
Since the map $\sigma(X,Y)=\textnormal{tr}(XY)$ is nondegenerate, it has all of the properties of a Frobenius form.
\end{squarematrix}



\begin{thebibliography}{9}

	\bibitem{Mac Lane 1988} Mac Lane, Saunders. {\em Categories for the Working Mathematician}. New York: Springer-Verlag, 1988.

  \bibitem{Kock 2003} Kock, Joachim. {\em Frobenius Algebras and 2D Topological Quantum Field Theories}. Cambridge: Cambridge University Press, 2003.

\end{thebibliography}

\end{document}
