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If A is a commutative algebra and B is an X- algebra, then the tensor product $A \otimes B$ is an X-algebra (so for example, $Com \otimes Lie$ is a Lie algebra). This is seen using the language of operads. Let $Com$ be the commutative operad. Since $Com(n)$ is a one dimensional vector space for every $n$, tensoring $Com$ with an operad $O$ doesn't change the operad $O$.

Does a similar thing hold true for a $C_\infty$ algebra? That is, if $A$ is a $C_\infty$ algebra, is $A \otimes B$ an $X_\infty$ algebra?

I'm still trying to familiarize myself with the language of operads, and perhaps the question can be made more precise in that language, where the infinity version of an operad is cofibrant resolution of the operad.

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I think your language is a bit mixed up. For example, I think the first sentence should read "If A is a commutative algebra and B is an X-algebra for some operad X, then A tensor B is an X-algebra"? –  Kevin H. Lin Aug 11 '10 at 20:26
you're right. its been edited –  Micah Miller Aug 11 '10 at 21:38

1 Answer 1

up vote 5 down vote accepted

We can see that this is false by looking at a degenerate example. Consider any operad $P$ so that $P(n)=0$ for $n>1$. If you tensor such an operad with $C_\infty$ so that $(P\otimes C_\infty)(n)=P(n)\otimes C_\infty(n)$ you get $P\otimes C_\infty=P$, since $C_\infty(1)$ is one dimensional and $P(n)=0$ for $n>1$. But $P$ is not the same thing as $P_\infty$ in general.

For a specific example, let $P(1)=k[x]/x^2$. Then $P_\infty(1)$ can be modeled by $(k[x_1,x_2,\cdots,],dx_k=\sum_{i+j=k}x_ix_j)$.

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