# Geometric treatment of the Ward-Takahashi identity

The quantum field theory generalisation of Noether's theorem about symmetries and conservation laws is the Ward-Takahashi identity. What is a suitable treatment of this in the context of differential geometry and a modern setting? Thanks

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Could you explain what you mean by Ward-Takahashi identity in your question, please? – user2013 Feb 27 '13 at 1:25
Your questions tend to be too terse in my opinion. Some context would be helpful. Why in the context of differential geometry? Modern from what point of view? Many symmetries of classical field theory Lagrangians do not survive in the quantum theory. There are anomalies, aka anomalous Ward-Takahashi identities. This is actually a huge subject so it is hard to answer without knowing in more detail what you are actually interested in. – Jeff Harvey Mar 13 '13 at 1:01

From a path integral point of view, the Ward-Takahashi identity is a straightforward consequence of the fundamental theorem of calculus.

Let $\delta$ be a vector field on the space $\mathcal{F}$ of fields which expresses the action of the group of gauge symmetries. Suppose that $d\phi$ is a measure which is invariant under these gauge symmetries. The fundamental theorem of calculus tells us that the integral of the total derivative

$\int_{\mathcal{F}} \delta(g) d\phi = 0,$

for any $g$ such that the contribution from the boundary of $\mathcal{F}$ is $0$.

In the special case where $g$ is the product $\mathcal{O}e^{-S}$ of a gauge-invariant observable and the exponential of a gauge-invariant action, we get a constraint on the path integral (expectation with respect to $e^{-S(\phi)}d\phi$):

$0 = \int_{\mathcal{F}} \delta(\mathcal{O}(\phi) e^{-S(\phi)})d\phi = \int_{\mathcal{F}}$\delta(\mathcal{O}) - \mathcal{O} \delta(S)$e^{-S(\phi)}d\phi = \langle \delta \mathcal{O}\rangle - \langle \mathcal{O} \delta(S) \rangle.$

If we use the fact that the variation of the action is the spacetime integral of the divergence of the current $\delta(S) = \int_\Sigma \nabla \dot{} J$, then we get the usual Ward-Takahashi identity:

$\langle \delta \mathcal{O}\rangle = \int_\Sigma \langle \mathcal{O} \nabla \dot{} J\rangle.$

Everything I've written here is only obviously true for finite-dimensional integrals. However, any path integral should be arbitrarily well approximated by such finite-dimensional integrals, so one can at least hope to transport the final result, even if the steps themselves can not be carried over.

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In many cases no invariant measure exists. There are anomalies. The passage from finite dimensional integrals to path integrals is more subtle than you indicate here. – Jeff Harvey Mar 13 '13 at 1:03
@JeffHarvey All true. I've only sketched out the basic idea above. The OP didn't write enough for me to bother with more than that. – userN Mar 13 '13 at 12:30