most recent 30 from http://mathoverflow.net 2013-05-22T02:15:24Z http://mathoverflow.net/feeds/question/35270 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/35270/extending-a-property-of-commutative-algebras-to-c-infinity-algebras Micah Miller 2010-08-11T20:14:28Z 2010-10-29T23:43:06Z

<p>If A is a commutative algebra and B is an X- algebra, then the tesnor 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.</p> <p>Does a similar thing hold true for a C-infinity algebra? That is, if A is a $C_\infty$ algebra, is $A \otimes B$ an $X_\infty$ algebra? </p> <p>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. </p>

http://mathoverflow.net/questions/35270/extending-a-property-of-commutative-algebras-to-c-infinity-algebras/44191#44191 Gabriel C. Drummond-Cole 2010-10-29T23:43:06Z 2010-10-29T23:43:06Z

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