Studying some Hopf Algebra in the context of mathematical physics. Looking at it as a generalization/extension of Tensor Products, Lie Algebra or Clifford Algebra.
Despite this being my n-th attempt, I fail to understand what a co-product actually does or means. Sure, it somehow complements the product, but in which sense?
Any hints to examples where it is actually possible to follow what's going on?
Any simple explanation/insight I am missing here?
On
Should have asked myself: "Have I googled it?" Easy game. A quick search for 'Hopf Algbra' and 'practical' lead to the brilliant Hopf Algebras in General and in Combinatorial Physics: a practical introduction
A real-world application of the coproduct for describing 'algebraic' composites. This almost exactly matches the hunch I had.
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This might sound dumb at first, but you should really view the coproduct as the dual of an 'ordinary' multiplication in the following sense: If $A\otimes_k A\to A$ is an associative multiplication over a finite-dimensional $k$-vector space $A$ , it dualizes to a map $\Delta: A^{\ast}\to (A\otimes_k A)^{\ast}\stackrel{\cong}{\longleftarrow} A^{\ast}\otimes_k A^{\ast}$, where the last isomorphism crucially relies on the finite-dimensionality of $A$.
How does that look like concretely? Assume for simplicity that $A$ has a multiplicative basis $\{e_i\}$, that is, $e_i e_j = e_k$ or $e_i e_j=0$, for some suitable $k$ depending on $i,j$. Then you can check that $\Delta$ maps $e_i^{\ast}$ to $\sum_{e_j e_k = e_i} e_j^{\ast}\otimes e_k^{\ast}$. In other words: The comultiplication describes all ways to decompose a (basis) element as a product of (basis) elements.
You can look at $A=k[x]/(x^n)$ for example and you'll see that $\Delta(x^k) = \sum_{a+b=k} x^a\otimes x^b$ corresponding to the decompositions $x^k = x^{a+b} = x^a x^b$. Note that the same example works with the full polynomial ring $A=k[x]$, when treating it as a graded vector space and using graded duals - in this case, you only need degree-wise finite-dimensionality.
But you see the issue with finite-dimensionality if you try something like $A=k[x^{\pm 1}]$: You have infinitely many decompositions of a single element, e.g. $1 = x x^{-1} = x^2 x^{-2} = ...$. Capturing all these would require an infinite sum of elementary tensors, which isn't defined in the ordinary tensor product.
Finally, note that if you have a comultiplication, you can in fact always dualize it to a multiplication, since the map $A^{\ast}\otimes_k A^{\ast}\to (A\otimes_k A)^{\ast}$ always exists. In that sense, you can see a comultiplication as a more restricted, 'finite' notion of a multiplication.