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May 5, 2017 at 21:10 answer added Vladimir Dotsenko timeline score: 4
May 5, 2017 at 20:52 comment added მამუკა ჯიბლაძე Your $t$, if exists, is uniquely determined by $t(1,x)=t(x,1)=\varepsilon(x)$, $t(x,x^2)=t(\Delta(x),x\otimes x)$, $t(x^2,x)=t(x\otimes x,\Delta(x))$, $t(x,x^3)=t(\Delta(x),x\otimes x^2)=t(\Delta(x),x^2\otimes x)$, etc. You just have to check whether what you obtain in this way is consistent or not.
May 5, 2017 at 20:42 history edited Ben McKay CC BY-SA 3.0
corrected the notation and the question
May 5, 2017 at 20:13 comment added unknown However, to avoid confusion the primitive property is not needed here.
May 5, 2017 at 20:12 comment added unknown There are plenty of bilinear maps $t:U\otimes U \to k$ that satisfy $t(x,x)=1$ and are not bialgebra pairings. For example on a basis, $t(x^n,x^m)=1 if m=n=1$, and zero otherwise. The point is that there exists a (unique) pairing $t$ which is a bialgebra pairing and fulfills $t(x,x)=1$. I want to know how to define the pairing.
May 5, 2017 at 20:00 comment added unknown Sorry it is typo it is all the same field $k$ through out. For example let $U=k[x]$ with the Hopf algebra structure determined by $x \in P(U)$. I mean the space $P(U)$ gives the coalgebra structure on the polynomial ring.
May 5, 2017 at 19:53 comment added Ben McKay I guess that $K$ is a field, but what is $k$? Is $K$ an extension field of $k$, perhaps?
May 5, 2017 at 19:52 comment added Ben McKay When you write $x \in P(K[x])$, do you mean two different things by the same letter $x$, one on the right hand side and the other one the left hand side?
May 5, 2017 at 19:42 comment added unknown &P(K[x])$ is the Space of primitive elements.
May 5, 2017 at 19:04 comment added Ben McKay What is $P(K[x])$?
May 5, 2017 at 19:03 history edited Ben McKay CC BY-SA 3.0
spelling, grammar, formatting
May 5, 2017 at 18:58 review First posts
May 5, 2017 at 19:00
May 5, 2017 at 18:56 history asked unknown CC BY-SA 3.0