Mathematics Asked by HaKuNa MaTaTa on December 3, 2020

An algebra is given by a triple $(A,mu,nu)$ where $A$ is a vector field over $k$, and

begin{gather*}

mu colon A otimes A rightarrow A, \

nu colon k rightarrow A

end{gather*}

are such that:

$mu$ is a linear map satisfying $mu circ (mu otimes mathrm{id}) = mu circ (mathrm{id} otimes mu)$; which is to say that $mu$ is associative.

The map $nu$ satisfies $mu circ (nu otimes mathrm{id}) = gamma$ where $gamma$ is the canonical isomorphism $gamma colon k otimes A rightarrow A$; this shows that $nu(1)$ is a unit in $A$

Can somebody explain to me how the constraint on $nu$ is equivalent to saying that $nu(1)$ is a unit in $A$? Thanks!

Both $mu circ ( nu otimes mathrm{id} )$ and $gamma$ are linear maps from $k otimes A$ to $A$. These maps coincide if and only if they are equal on simple tensors, i.e. we have $mu circ (eta otimes mathrm{id}) = gamma$ if and only if $$ ( mu circ (nu otimes mathrm{id}) )(lambda otimes a) = gamma(lambda otimes a) $$ for all $lambda in k$ and $a in A$. In other words, we need that $$ nu(lambda) cdot a = lambda a $$ for all $lambda in k$ and $a in A$. Here we denote on the left hand side by $cdot$ the multiplication on $A$ coming from $mu$, and on the right hand side we have the scalar multiplication coming from the vector space structure of $A$. By the linearity and $nu$ and the bilinearity of the multiplication $mu$ we can rewritte the left hand side of this required equation as $$ nu(lambda) cdot a = nu(lambda 1) cdot a = ( lambda nu(1) ) cdot a = lambda ( nu(1) cdot a ) ,. $$ We can hence rewrite the above equality as $$ lambda ( nu(1) cdot a ) = lambda a $$ for all $lambda in k$ and $a in A$. We see that this holds for all $lambda$, $a$ if and only if it holds for $lambda = 1$ and all $a$. In other words, the identity $mu circ (eta otimes mathrm{id}) = gamma$ holds if and only if $$ nu(1) cdot a = a $$ for all $a in A$. But this condition means precisely that $nu(1)$ is a multiplicative neutral element of $A$ with respect to $mu$.

Correct answer by Jendrik Stelzner on December 3, 2020

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