Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I'm reading Hatcher's book on algebraic topology. In Section 3.C, he proves as Theorem 3C.4 that if $A$ is a graded commutative associative Hopf algebra over a field of characteristic $0$ and $A^n$ is finitely generated for each $n$, then $A$ is isomorphic as an algebra to the tensor product of an exterior algebra on odd-dimensional generators and a polynomial algebra on even-dimensional generators. I emphasized the phrase as an algebra because I first assumed that this was a typo on his part, but indeed in his proof he only shows that $A$ has the right algebra structure. The coproduct is used to establish this, but he does not prove that $A$ has the same coproduct as the indicated tensor product.

Question : Must it? In other words, over a field of characteristic $0$ can the tensor product of an exterior algebra on odd-dimensional generators and a polynomial algebra on even-dimensional generators be given a Hopf-algebra structure other than the one coming from the tensor product?

share|improve this question
    
I think you should put "connected" in your statement or else it fails. –  darij grinberg Jan 21 '13 at 17:51
1  
Anyway, it's easier to find counterexamples to the dual statement: Every locally finite graded connected cocommutative associative Hopf algebra over a field of characteristic $0$ is isomorphic to a symmetric coalgebra (in the graded sense, so really a tensor product of an even symmetric and an odd exterior algebra) as a coalgebra, but not as an algebra. For instance, when $\mathfrak g$ is a Lie algebra, then the symmetrization map $S\left(\mathfrak g\right) \to U\left(\mathfrak g\right)$ (or was it the other way round?) is a coalgebra isomorphism, but the two algebras are (generally) ... –  darij grinberg Jan 21 '13 at 17:53
    
... not isomorphic (one of them is commutative, the other isn't). By taking graded duals, this becomes a counterexample to your question. –  darij grinberg Jan 21 '13 at 17:53

2 Answers 2

up vote 1 down vote accepted

There are generalizations of the following construction, but here is a simple one. Let $\mathfrak n$ denote the Lie algebra of strictly-upper-triangular $n\times n$ matrices. It is $\mathbb Z_{>0}$-graded, by declaring that for $1\leq k \leq n$, the degree-$k$ part of $\mathfrak n$ consists of those matrices supported on the diagonal that is $k$ steps above the main diagonal. (So $\mathfrak n_k$ has a basis consisting of those matrices with a $1$ in the $(i,i+k)$th spot and $0$s elsewhere.) This is a bosonic grading, so to match Hatcher's conventions, you perhaps should double the grading.

In any case, consider the polynomial algebra $\operatorname{Sym}(\mathfrak n^\ast)$ generated by the dual vector space — so this is the algebra $\mathcal{O}(\mathfrak n)$ of polynomial functions on $\mathfrak n$. It is connected and $\mathbb Z_{\geq 0}$-graded. It has many compatible coproducts. One is the canonical coproduct generated by $x \mapsto x\otimes 1 + 1\otimes x$ for $x\in \mathfrak n^\ast$. Another is the Baker–Campbell–Hausdorff formula $\operatorname{BCH}$. In detail, let $N$ denote the group of matrices which have $1$s on the diagonal and $0$s below it, and let $\operatorname{m} : N\times N \to N$ denote the map of matrix multiplication. Then the matrix $\exp : \mathfrak{n} \to N$ is a degree-$n$ polynomial, as is its inverse $\log: N \to \mathfrak{n}$; we can therefore pull back polynomial functions along these, so for example we have $\operatorname{m}^\ast : \mathcal{O}(N) \to \mathcal{O}(N)\otimes \mathcal{O}(N)$. Then: $$\operatorname{BCH} = (\exp^\ast \otimes \exp^\ast) \circ \operatorname{m}^\ast \circ \log^\ast.$$ It is a standard fact that $\operatorname{BCH}$ preserves the grading on $\operatorname{Sym}(\mathfrak n^\ast)$, and makes $\operatorname{Sym}(\mathfrak n^\ast)$ into a Hopf algebra.

share|improve this answer
    
A different way of saying this is that the affine space $k^n$ is a Lie group in many different ways (just pick an nilpotent Lie algebra structure on it and «exponentiate» it) Each such structure puts a different Hopf algebra structure on of $\mathcal O(k^n)=k[x_1,\dots,x_n]$. –  Mariano Suárez-Alvarez Jan 21 '13 at 17:23
    
(And all of these will be connected, generated in degree one and with all its generators primitive (they can only be primitive, for degree reasons) so they satisfy the extra conditions imposed by Roger in comment to PB's answer) –  Mariano Suárez-Alvarez Jan 21 '13 at 17:26
    
PB's answer came first, but I found this one to be more enlightening, so I accepted it. Thanks to both of them! –  Roger Jan 21 '13 at 17:32

To clarify the Theorem I would say, ".... If A is a graded commutative Hopf algebra then, the underlying algebra structure is tensor of exterior and a polynomial." And on the same underlying algebra, one can have two different coproduct

Let $Q[x]$ be polynomial of algebra where $|x| =0$. A coproduct will be a map of $Q[x]$-algebra, and its enough to define the map on $|x|$. So I will describe them as follows

$\psi_{1}(x)= x \otimes 1 + 1 \otimes x$

$\psi_{2}(x)= x \otimes x$.

share|improve this answer
    
Is it possible to construct examples that are connected (so the degree $0$ part is just the field) and where (as Hatcher requires in his definition, though I understand that it is not always required) the coproduct of $x$ is $x \otimes 1 + 1 \otimes x + \sum_{i=1}^{|x|-1} y_i \otimes y_{|x|-i}'$ with $|y_j|=|y_j''| = j$? –  Roger Jan 21 '13 at 16:47
    
Roger, it is best if you add these extra conditions to the question itself. –  Mariano Suárez-Alvarez Jan 21 '13 at 17:06
2  
Take two generators $x$ and $y$ of degree $2$ and $4$, respectively, and define $\Delta(x)=x\otimes1+1\otimes x$ and $\Delta(y)=y\otimes 1+\lambda x\otimes x+1\otimes y$ for $\lambda$ in the basefield. –  Mariano Suárez-Alvarez Jan 21 '13 at 17:19
    
This is a very nice example. –  Prasit Jan 21 '13 at 22:14

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.