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I'm working through Kassel's Quantum Groups book and I'm stuck on what looks like a pretty simple problem (ch3, ex 2 in the book):

Let $k$ be a field and $C= k[t]$ be a coalgebra with coproduct: $$ \Delta (t^n) = \sum_{p+q = n} t^p \otimes t^q$$

and product

$$ t^n \cdot t^m = \binom{n+m}{n} t^{n+m} $$

The book asks me to prove that this, together with the co/unit, form a bialgebra. However, if I try to calculate $\Delta(x)\Delta(y)$ with that product I get:

$$ \Delta(t^n)\Delta(t^m) = \Large( \sum_{i+j=n} t^i\otimes t^j \Large) \Large( \sum_{k+l=m} t^k\otimes t^l \Large)$$ $$ = \sum_{i+j+k+l=n+m} \binom{i+k}{i}\binom{j+l}{j} t^{i+k} \otimes t^{j+l}$$

Which does equal $\Delta(t^n \cdot t^m) = \binom{n+m}{n}\sum_{p+q=n+m} t^p\otimes t^q$

However, if I ignore Kassel's product with the binom coeff, then I have bialgebra without issue:

$$ \Delta(t^n)\Delta(t^m) = \Large( \sum_{i+j=n} t^i\otimes t^j \Large) \Large( \sum_{k+l=m} t^k\otimes t^l \Large)$$ $$ = \sum_{i+j+k+l=n+m} t^{i+k} \otimes t^{j+l} =\Delta(t^{n+m})$$

Clearly the book contains an error or I don't understand something. Probably the latter. Can you help me figure out what?

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3  
Your second summation runs over the wrong index set. It's still $\sum_{i+j = n, k+l = m}$. What you've written is not equivalent. –  Qiaochu Yuan Aug 1 '12 at 23:20
4  
You have some other algebra errors as well. I would suggest writing out all your expressions in full, without $\Sigma$'s, for some small values of $(m,n)$, such as $(2,3)$. –  David Speyer Aug 1 '12 at 23:24

1 Answer 1

up vote 0 down vote accepted

As the commentators have pointed out, I can't do basic algebra! Let's try this again a bit more carefully....

$$ \Delta(t^n)\Delta(t^m) = \Large( \sum_{i+j=n} t^i\otimes t^j \Large) \Large( \sum_{k+l=m} t^k\otimes t^l \Large)$$ $$ = \sum_{i+j=n, k+l=m} t^i \cdot t^k \otimes t^j \cdot t^l = \sum_{j=0}^m \sum_{k=0}^n t^{m-j} \cdot t^{n-k} \otimes t^{j} \cdot t^{k}$$ $$ \sum_{j=0}^m \sum_{k=0}^n \binom{m-j+n-k}{m-j}\binom{j+k}{j} t^{m-j+n-k} \otimes t^{j+k} $$

From this we can write out an expression for the coeff of $t^{m+n-s}\otimes t^s$ for each s in the above sum:

$$\sum_{p+q=s} \binom{m-p+n-q}{m-p}\binom{p+q}{p} = \sum_{p=0}^s \binom{m+n-s}{m-s+p}\binom{s}{s-p} = \binom{m+n}{m}$$ as desired. Correct?

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