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Why is an extended T(Q)FT called fully local?

+ 3 like - 0 dislike
63 views

Hopefully this question does not double another. If so, don't bother to close this.

An extended topological quantum field theory is sometimes called, 'fully local". Why is that? I can imagine that such a theory Has local structure, while an ordinary TQFT has not.

The question is vague, but a more precise answer, or even an example of local stuff going on, would be appreciated.


This post imported from StackExchange MathOverflow at 2015-06-10 18:55 (UTC), posted by SE-user Mark.Neuhaus

asked May 12, 2015 in Theoretical Physics by mark.neuhaus (15 points) [ revision history ]
edited Jun 10, 2015 by Dilaton

1 Answer

+ 4 like - 0 dislike

I believe there's an explanation on the nlab page for extended topological quantum field theories (http://ncatlab.org/nlab/show/extended+topological+quantum+field+theory) and on page 13 of Lurie's cobordism paper. We can compute the value of an extended TQFT $Z$ on a manifold $X$ simply by computing it locally, and then gluing. In fact, we can determine a $1$-dimensional TQFT by its value on the point!

For example, let us return to the case where we wished to compute the value of an extended TQFT $Z$ on a manifold $X$. Locally we can break $X$ down into a collection of neighborhoods of points; if we wished to compute $X$, we could simply study the value of $Z$ on these neighborhoods! Another example is when a manifold $X$ is given by the data of a cobordism, say $X=X_1\coprod_{\partial X_1=\partial X_2}X_2$. If $Z$ is a $1$-dimensional TQFT, then $Z(X)$ is determined by the value of $Z$ on $X_1$ and $X_2$.

I don't know much about quantum field theories from a physics perspective, but the nlab claims that "the notion of locality in quantum field theory is precisely this notion of locality. And, as also discussed at FQFT, this higher dimensional version of locality is naturally encoded in terms of n-functoriality of $Z$ regarded as a functor on a higher category of cobordisms".

In a single sentence (which I've already written at the beginning of the answer): we can compute the value of an extended TQFT $Z$ on a manifold $X$ simply by computing it locally, and then gluing.

This post imported from StackExchange MathOverflow at 2015-06-10 18:55 (UTC), posted by SE-user Sanath
answered May 12, 2015 by Sanath (40 points) [ no revision ]

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