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Let $k$ be a field and let $(C, \Delta, \epsilon)$ be a vector space which is a coalgebra. Let $(M, \delta)$ be a comodule. Suppose $V \subseteq M$ is a finite dimensional space. For each $v \in V$, we can write

$ \delta(v) = \sum_{i < \lambda}u_i(v) \otimes c_i(v) $ where the cardinal $\lambda= \dim V \dim C$ and cofinitely many of the $u_i (v)$ are $0$. Let $\{ v_i \colon i = 1, \ldots, n\}$ be a basis of $V$. Let $N$ be the vector space with basis $\{ u_j (v_i) \mid i = 1, \ldots, n ; j< \lambda \}$. This is finite dimensional.

I am trying to show that $\delta(N) \subseteq N \otimes C$. This would amount to showing that $\delta(u_j(v_i)) \in N \otimes C $ for every $j,i$.

Here is what I understand so far: we have $\delta(v_i) = \displaystyle\sum_{j < \lambda}u_j(v_i) \otimes c_j(v_i)$ and if we apply $\delta \otimes 1$ to this we get $ \sum_{j < \lambda}\delta(u_j (v_i) )\otimes c_j (v_i) = \sum_{j < \lambda}u_j (v_i) \otimes \Delta (c_j (v_i )) $

It seems that with this I am close but I am not sure how to show that $\delta (u_j (v_i))$ actually lies in $N \otimes C$.

Any help would be appreciated.

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    I guess you don't need to parametrise the $c_i$'s by $v$, we can just say \delta(v) = \sum_{i< \lambda} u_i(v) \otimes c_i where \{ c_i \mid i < \lambda \} is a basis for $C$. Thanks I will give this some thought.2012-10-24

0 Answers 0