I need your help on how to show the existence of a bialgebra pairing: for the polynomial ring $k[x]$ over a field $k$, there is a bialgebra pairing $t:k[x]\otimes k[x]→k$ such that $t(x,x)=1$. What is the unique bialgebra pairing satisfying $t(x,x)=1$?
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4$\begingroup$ What is $P(K[x])$? $\endgroup$– Ben McKayCommented May 5, 2017 at 19:04
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$\begingroup$ &P(K[x])$ is the Space of primitive elements. $\endgroup$– unknownCommented May 5, 2017 at 19:42
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1$\begingroup$ 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? $\endgroup$– Ben McKayCommented May 5, 2017 at 19:52
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$\begingroup$ I guess that $K$ is a field, but what is $k$? Is $K$ an extension field of $k$, perhaps? $\endgroup$– Ben McKayCommented May 5, 2017 at 19:53
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$\begingroup$ 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. $\endgroup$– unknownCommented May 5, 2017 at 20:00
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1 Answer
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The bialgebra pairing condition implies $(1,b_1b_2)=(\Delta(1),b_1\otimes b_2)=(1,b_1)(1,b_2)$, so in particular $(1,1)=(1,1)^2$, and $(1,1)$ is zero or one. Also by a similar calculation, $(b_1b_2,1)=(b_1,1)(b_2,1)$.
We have $$(x,1)=(x,1\cdot1)=(\Delta(x),1\otimes 1)=(x\otimes 1+1\otimes x,1\otimes1)=(x,1)(1,1)+(1,x)(1,1).$$ If $(1,1)=0$ this implies that $(x,1)=0$, and if $(1,1)=1$, this implies that $(1,x)=0$. Together with a similar evaluation of $(1,x)$ this means that $(x,1)=(1,x)=0$. Since $(1,b_1b_2)=(1,b_1)(1,b_2)$ and $(b_1b_2,1)=(b_1,1)(b_2,1)$, we have $(1,x^n)=(x^n,1)=0$ for $n>0$.
We have $$ 1=(x,x)=(x,x\cdot 1)=(\Delta(x),x\otimes 1)=(x\otimes 1+1\otimes x,x\otimes 1)=(x,x)(1,1)+(1,x)(x,1)=(1,1), $$ so $(1,1)=1$.
Clearly, we have $$ (x^m,x^n)=(\Delta(x^m),x\cdot x^{n-1})=\left(\sum_{j=0}^m\binom{m}{j}x^j\otimes x^{m-j},x\otimes x^{n-1}\right). $$
Let us show by induction that $(x^m,x^n)=0$ for $k\ne n$. We already know it for $(m,n)=(0,1),(1,0)$. Suppose it is established for $m+n<N$, where $N\ge 2$. Let us consider $m+n=N$. We may assume $m,n>0$ since $(1,x^k)=(x^k,1)=0$. By the formula above we have $$ (x^m,x^n)=\sum_{j=0}^m\binom{m}{j}(x^j\otimes x^{m-j},x\otimes x^{n-1})=\sum_{j=0}^m\binom{m}{j}(x^j,x)(x^{m-j},x^{n-1}). $$ If $j=0$ or $j=m$, we have $(x^j,x)(x^{m-j},x^{n-1})=0$ since $(1,x^k)=(x^k,1)=0$. Suppose that $0<j<m$. In this case $j+1<m+n$, hence if $j\ne 1$ we have $(x^j,x)=0$ by induction, and if $j=1$, we have $(x^{m-1},x^{n-1})=0$ by induction as $m-1+n-1<m+n$.
Finally, by the same formula for $m=n$ $$ (x^n,x^n)=\sum_{j=0}^n\binom{n}{j}(x^j\otimes x^{m-j},x\otimes x^{n-1})=\sum_{j=0}^n\binom{n}{j}(x^j,x)(x^{n-j},x^{n-1}), $$ and by the result we just proved only $j=1$ has a non-zero contribution, so $$ (x^n,x^n)=n(x^{n-1},x^{n-1}), $$ which immediately gives $$ (x^n,x^n)=n! $$
answered May 5, 2017 at 21:10
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$\begingroup$ The answer is perfect for the previous version of the question. Note that now it is not presumed anymore that $x$ is primitive. $\endgroup$ Commented May 5, 2017 at 21:34
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$\begingroup$ @მამუკაჯიბლაძე : I assume that the author of the question means the standard bialgebra structure on the polynomial ring when writing "the polynomial ring". I still think that "the primitive property is not needed here" means that the original version of the question was overloaded by unnecessary notation, not that $x$ is not assumed primitive. $\endgroup$ Commented May 5, 2017 at 22:04
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$\begingroup$ Thanks. That's really helpful. Also can I use the primitive properties to say that $t$ is non-degenerate if char $k=0$?. $\endgroup$– unknownCommented May 5, 2017 at 22:32
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$\begingroup$ @UmarMuhammadFaruq this calculation does indeed show that the pairing is nondegenerate in char 0. It assumes the "standard" bialgebra structure on $k[x]$ for which $x$ is a primitive element, and nothing else. $\endgroup$ Commented May 5, 2017 at 22:49
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1$\begingroup$ @UmarMuhammadFaruq If the ground field is of characteristic $p$, then the kernel of the pairing is the ideal generated by $x^p$, since $n!$ is divisible by $p$ for all $n\ge p$. $\endgroup$ Commented May 5, 2017 at 23:08