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Let $C$ be a coalgebra and $\Delta: C \to C\otimes C$ a co-multiplication map. Then, due the co-associative property we can consider $\Delta^m$. But how is defined $\Delta^{m}: C \to C^{\otimes m}$?

Given $f,g \in C$ and $1\leq k \leq m$ can we have

$$\begin{align*}\Delta^{m-1}(fg)&=\Delta^{k-1}(fg) \otimes \operatorname{id}^{m-k} + \Delta^{k-1}(f)\otimes \Delta^{m-k-1}(g) \\ &+\Delta^{k-1}(g)\otimes \Delta^{m-k-1}(f)+\operatorname{id}^{\otimes k} \otimes \Delta^{m-k-1}(fg)?\end{align*}$$

Thanks.

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    $\begingroup$ One question mark is enough in most circumstances! $\endgroup$
    – Mariano Suárez-Álvarez
    Oct 24, 2010 at 14:50
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    $\begingroup$ But not so with exclamation marks, since 5! = 120 but 5!! = 15. $\endgroup$
    – Michael Joyce
    Jun 30, 2020 at 4:57

3 Answers 3

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To add to the answer above, I'd like to advertise the so-called sumless Sweedler notation here.

This notation works as follows: let $C$ be a coalgebra; then if $c \in C$ then we write $\Delta(c) = c_1 \otimes c_2$ as an abbreviation of the more precise $\Delta(c) = \sum_{i=1}^k c_{i1} \otimes c_{i2}$ for some $c_{i1}, c_{i2} \in C, i = 1,\ldots, k$. Then if we wish to consider $(1 \otimes \Delta) \circ \Delta : C \to C\otimes C \otimes C$ we simply write

$$[(1 \otimes \Delta) \circ \Delta] (c) = c_1 \otimes c_2 \otimes c_3.$$

In this notation, the coassociativity axiom $(1 \otimes \Delta) \circ \Delta = (\Delta \otimes 1) \circ \Delta$ becomes the inevitable

$$ c_1 \otimes (c_2 \otimes c_3) = (c_1 \otimes c_2) \otimes c_3.$$

Sweedler's book "Hopf algebras", Susan Montgomery's book "Hopf algebras and their actions on rings" and the Wikipedia article http://en.wikipedia.org/wiki/Coalgebra are good references.

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    $\begingroup$ Indeed, the exercises in Sweedler's book regarding the notation are essentially mandatory for anyone trying to work with these things! $\endgroup$
    – Mariano Suárez-Álvarez
    Feb 28, 2011 at 16:35
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    $\begingroup$ Actually, coassociativity is in terms of the Sweedler notation $c_1\otimes(c_2)_1\otimes(c_2)_2=(c_1)_1\otimes (c_1)_2\otimes c_2$. $\endgroup$
    – Mariano Suárez-Álvarez
    Jan 15, 2017 at 18:57
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We usually define $$\Delta^3=(\mathrm{id}_C\otimes\Delta)\circ\Delta,$$ and, more generally, $$\Delta^{k+1}=(\mathrm{id}_C^{k-1}\otimes\Delta)\circ\Delta^k.$$

(Your last formula does not make sense in a coalgebra: you seem to be multiplying elements of $C$; even in bialgebra, though, things like "$\Delta(fg)\otimes\mathrm{id}$" are pretty strange...)

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    $\begingroup$ Both equations are not consistent! Anyway, have you meant $\Delta^2$ in the first formula or the power is wrong in the second one? $\endgroup$
    – Binai
    Oct 24, 2010 at 17:34
  • $\begingroup$ One more point. Observe that the usual action of $fg$ on $u\otimes v$, where $f,g$ are two elements in the Universal Enveloping Algebra $U(G)$ of a Lie algebra $G$ and $u,v$ are elements of a representation $V$ of $G$, is given by $fg (u \otimes v)=fgu \otimes v + fu \otimes gv + gu\otimes fv + u\otimes fgv$ using the comultiplication, right? How to state this fact for $V^{\otimes n}$, i.e. $fg$ acting on $u \otimes v$, where $u=\otimes_{i=1}^{n-k} u_i$ and $v=\otimes_{i=1}^k v_i$? Thanks, $\endgroup$
    – Binai
    Oct 24, 2010 at 17:55
  • $\begingroup$ if you look at your original definition chris you will see that $\Delta^2 = \Delta$. $\endgroup$
    – Sean Tilson
    Oct 24, 2010 at 17:59
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    $\begingroup$ @Chris: The numbering convention is for $\Delta^n$ to denote the map $C \to C^{\otimes n}$ --- it has precisely $n$ outputs. You can extend this backwards: $\Delta^2 = \Delta$, $\Delta^1 = \operatorname{id}$, and, if your coalgebra is counital, $\Delta^0$ is the counit. $\endgroup$
    – Theo Johnson-Freyd
    Oct 24, 2010 at 19:00
  • $\begingroup$ And in a bialgebra then the comultiplication is an algebra homomorphism so one would have $\Delta(f g) = \Delta(f) \Delta(g)$ and so on for the iterates. $\endgroup$
    – Andrew Stacey
    Oct 25, 2010 at 7:47
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When working with coalgebras I often find it helpful to dualize and to consider the corresponding situation for multiplication $\mu: A \otimes A \to A$ with an algebra $A$.

In this point of view $\Delta^m$ corresponds to $\mu^m: A^{m+1} \to A$, given by $$\mu^m(a_0 \otimes ... \otimes a_m) = (a_0 ... a_{m-1}) a_m.$$ It's easy to see that this formula is equivalent to $$\mu^m = \mu \circ (\mu^{m-1} \otimes id_A).$$ Dualizing again yields $$\Delta^m = (\Delta^{m-1} \otimes id_C) \circ \Delta.$$ BTW: If $\Delta$ is coassociative this equals Mariano's formula $$\Delta^m = (id_C^{\otimes m-1} \otimes \Delta) \circ \Delta^{m-1}.$$

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