19
$\begingroup$

Can someone please explain to me what Euler Density is? I have encountered it in Weyl anomaly related issues in various articles. Most of them assumes that its familiar, but I couldn't find any accessible paper or a book discussing that. So, it would be nice if I can understand what it is physically and mathematically and also find a reference where I can look it up.

Also related to that it would be nice to find a reference where people have derived $\langle T_i^i\rangle$ in curved background which involves Euler Density, $W^i$ etc.

$\endgroup$

1 Answer 1

20
$\begingroup$

Euler density is simply the integrand in $2n$ dimensions of the integral that is equal to the Euler characteristic. The Euler characteristic may be written as the integral of the following Euler density in $2n$ dimensions: $$E_{2n} = \frac{1}{2^n} R_{i_1 j_1 k_1 l_1} \dots R_{i_n j_n k_n l_n} \epsilon^{i_1 j_1 \dots i_n j_n} \epsilon^{k_1 l_1 \dots k_n l_n} $$ Note that for $n=1$ i.e. in two dimensions, it is linear in the Riemann tensor - and therefore also in the Ricci scalar (because the Riemann tensor is fully determined by the Ricci scalar in 2D). In four dimensions, the Euler density is quadratic in the Riemann tensor, and so on.

The Euler character - a "regularized number of points in a manifold" - may also be calculated in many other ways, e.g. for polytopes by adding the number of faces, subtracting edges, adding vertices, etc. For nice manifolds, it's only nonzero for even-dimensional manifolds. For closed orientable two-dimensional Riemann surfaces, it is given by $2-2h$ where $h$ is the number of handles (the genus also known as $g$). One may construct a general open/closed orientable/unorientable two-dimensional manifold by adding $b$ (circular) boundaries i.e. holes and $c$ crosscaps (holes with identified antipodal points, creating an unorientable manifold) and the total Euler characteristic is then $$ \chi = 2-2g - b - c.$$ You should imagine that if there is a function $L(\sigma^i)$ depending on the manifold's coordinates sigma such that $L$ has units of $U$, the path integral $\int DL(\sigma^i)$ has units of $U^\chi$ where $\chi$ is the Euler characteristic: that's what I meant by saying that $\chi$ is the regularized number of points.

So the Euler characteristic (or character) is arguably the most important and most elementary topological invariant of a manifold. The fact that the integral of $E_{2n}$ is a topological invariant may be seen by calculating its variation which vanishes (for any variation of the metric) - one reduces the derivative to some of the standard identities for the Riemann tensor, especially the two Bianchi identities involving antisymmetrization (and, in one case, one derivative).

The derivation of the trace of the stress-energy tensor is done for $d=2$ in Polchinski's "String Theory", Volume I. Equation (3.4.31) says $$T^a_a (\sigma) = -\frac{C}{12} R(\sigma)$$ where $R$ is the Ricci scalar, also interpretable as a multiple of the Euler density. The $C$ ends up being a definition of the central charge. I don't know the general form of a similar equation in $d$ dimensions but its exact form - at least the parameters - do depend on the theory. I guess that in general, the trace is equal to some linear combination of the Euler density and perhaps some other generators besides the stress-energy tensor.

$\endgroup$
4
  • $\begingroup$ Hi Lubos, thanks for your answer (as always). I was looking at hep-th/0404176 of Kostas Skenderis, and found Eq. 17. If you get a time to look at that, you will find the general expression of $T^i_i$. Then, I was wondering if there's any derivation of that. $\endgroup$
    – user1349
    Commented Mar 13, 2011 at 6:28
  • 1
    $\begingroup$ Hi, yes, there's a way how to derive it but even the result is pretty complicated so it's clear that the derivation won't be too much easier. It may always be derived in similar ways as in Chapter 3 of Polchinski's book - but Skenderis considers a slightly more general theory. You know, the form of the trace of the stress-energy tensor is very restricted because it must vanish on a flat metric background and when all the conservation laws etc. are obeyed. So the trace has to be a combination of curvature-like quantities of the metric and failures of conservation laws (divergences). $\endgroup$ Commented Mar 13, 2011 at 7:13
  • $\begingroup$ 9302047 is the article which first derives it. $\endgroup$
    – user1349
    Commented Aug 25, 2011 at 20:49
  • $\begingroup$ A small comment regarding the vanishing of the variation of the Euler density: in addition to the standard identities for the Riemann tensor, you'll also need dimensionally dependent identities. In e.g. two dimensions the variation yields the Einstein tensor, whose vanishing can be derived by taking appropriate contractions of $R_{i_1 i_2}{}^{[j_1 j_2} \delta_{i_3}^{j_3]}$. $\endgroup$ Commented Jul 4, 2013 at 19:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.