tensor product of Hopf algebras and coalgebraic structure on the dual algebra I am learning Hopf algebras, and there are two questions as follows:


*

*Is the tensor product of two Hopf algebras still a Hopf algebra?


*Let $A$ be an infinite dimensional algebra. Is the dual $A^*$ a coalgebra? (In the case of the dimension of $A$ is finite the answer is positive.)

If they are, where can I find the proofs?
Thanks for your help.
 A: Given two Hopf algebras $\big(H,\mu_H, \eta_H, \Delta_H, \varepsilon_H, S_H\big)$, $\big(G,\mu_G, \eta_G, \Delta_G,\varepsilon_G, S_G\big)$ over a field $k$, we can form the tensor product Hopf algebra $\big(H\otimes G,\mu_{H\otimes G}, \eta_{H\otimes G}, \Delta_{H\otimes G},\varepsilon_{H\otimes G},S_{H\otimes G}\big)$ through: 
$$\mu_{H\otimes G}=(\mu_H\otimes\mu_G)\circ(Id\otimes\tau\otimes Id) :H\otimes G \otimes H \otimes G \rightarrow H\otimes G
\\ \\ \\ 
\eta_{H\otimes G}=(\eta_H\otimes\eta_G)\circ\phi^{-1}:k\rightarrow H\otimes G
\\ \\ \\  
\Delta_{H\otimes G}=(Id\otimes\tau\otimes Id)\circ(\Delta_H\otimes\Delta_G):H\otimes G\rightarrow H\otimes G \otimes H \otimes G , \\ \\ \\
 \varepsilon_{H\otimes G}=\phi \circ (\varepsilon_H\otimes\varepsilon_G):H\otimes G\rightarrow k \\ \\ \\
S_{H\otimes G}=S_H\otimes S_G:H\otimes G\rightarrow H\otimes G
$$
where $Id$ is the identity map and $\phi:k\otimes k\stackrel{\cong}{\rightarrow} k$ the natural isomorphism. In other words, the standard tensor product of algebras is combined with the tensor product coalgebra structure and the antipode is given by the tensor product of antipodes. This is a standard construction and can be found in almost any introductory Hopf algebra text. 
Now, regarding your second question, as you correctly mention in your post, it holds if the algebra $(A,\mu,\eta)$ is finite dimensional: then $(A^*,\lambda^{-1}\circ\mu^*,\psi^{-1}\circ\eta^*)$ is a finite dimensional coalgebra. Here we denote with $\psi$, the natural v.s. isomorphism 
$k\cong k^*$ defined by $\psi(\kappa)=f$ with $f(1)=\kappa$ and $\psi^{-1}(f)=f(1)$ and with $\lambda:A^*\otimes A^*\rightarrow(A\otimes A)^*$ the linear v.s. injection, defined by $\lambda(f\otimes g)(u\otimes v)=f(u)g(v)\in k$. Not that, in case $A$ is of finite $k$-dimension then $\lambda$ is a vector space isomorphism. 
However, if $A$ is infinite dimensional, then the above construction meets the following problem: Since $\lambda$ is injective but not an isomorphism, $\lambda^{-1}$ is not well defined: $A^*\otimes A^*$ is a proper subspace of $(A\otimes A)^*$ and thus the image of $\mu^*:A^*\rightarrow(A\otimes A)^*$ may not lie inside $A^*\otimes A^*$. The problem is treated in the following manner: Instead of considering the whole of the (linear) dual space $A^*$, which seems too "big" to carry the construction, we consider its linear subspace $A^\circ$ defined as 
$$
A^\circ=\{f\in A^*\big|\ker f \textrm{ contains an ideal of finite codimension} \}=\{f\in A^*\big|f(I)=0 \textrm{ for some ideal } I \textrm{ of } A \textrm{ such that } \dim A/I<+\infty\} 
$$
$A^\circ$ and its elements  can be characterized by the equivalent conditions of the following proposition: 
Proposition: Let $A$ be a $k$-algebra and $f\in A^*$. Then the following conditions are equivalent:


*

*$f\in A^\circ$

*$\mu^*(f)\in\lambda(A^*\otimes A^*)$

*$\mu^*(f)\in\lambda(A^\circ\otimes A^\circ)$

*$A\rightharpoonup f$ is finite dimensional

*$f\leftharpoonup A$ is finite dimensional

*$A\rightharpoonup f\leftharpoonup A$ is finite dimensional


$A^\circ$ is called the restricted dual of $A$ and in fact it is the largest subspace $V$ of $A^*$ such that $\mu^*(V)\subseteq V\otimes V$. 
Now, it can be shown that $A^\circ$ together with the maps  $(\lambda^{-1}\circ\mu^*)$, $(\psi^{-1}\circ\eta^*)$ as before, is a $k$-coalgebra. More details on the proofs of this construction can be found in various sources. See for example, 


*

*Dascalescu, Nastasescu, Raianu, "Hopf algebras, an introduction"

*S. Montgomery, "Hopf algebras and their actions on rings"

*Moss E. Sweedler, "Hopf algebras"

*Kassel, "Quantum groups"
etc 

A: Related question: You can see darij grinberg's answer for 1).
The answer for 2) is negative. For example, you can see here

while the K-dual of every coalgebra is an algebra, the K-dual of an infinite-dimensional algebra need not be a coalgebra.

ON THE APPLICATIONS OF COALGEBRAS TO GROUP ALGEBRAS

Unfortunately, the dual of an infinite dimensional algebra is too large to canonically take a coalgebra structure. 

