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One loop corrections for gravity usually includes non-local terms in the action such as $R\log(\frac{-\Box}{\mu^2})R$, where $\Box=g^{\mu\nu}\nabla_\mu\nabla_\nu$ is the D'Alembert operator, $R$ is the scalar curvature and $\mu$ is just some constant. My problem concerns how to interpret the log of an operator. In http://arxiv.org/pdf/1507.07829.pdf, the definition goes like (see equation 4.2 in pg 9)

$ \log(\frac{-\Box}{\mu^2})=\int_0^\infty ds \left(\frac{1}{\mu^2+s}+\frac{1}{-\Box+s}\right). $

Besides this definition be quite arbitrary to me, again a function of $\Box$ shows up and I end up with the same problem. So how should I interpret a function like $f(\Box)$? Is it to be seen as a Taylor expansion in the same way that is usually defined in quantum mechanics, i.e.

$ f(\Box)=\sum_n\frac{f^{(n)}(0)}{n!}\Box^n \quad? $

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To promote a function $f(x)$ of a number to a function of a linear operator $f(A)$, one way to proceed is to diagonalize the operator $A = V^{-1} D V$ where $D$ is diagonal. We then let $f$ act on each of the eigenvalues (i.e. the diagonal entries of $D$) and define $f(A)$ to be $$ f(A) \equiv V^{-1} f(D) V. $$ Of course if $A$ is not diagonalizable, then things can be trickier. The Wikipedia entry for the log of a matrix discusses how to first put the operator in Jordan normal form and then how the Taylor series of $\log(I+K)$ naturally truncates for each Jordan block. Hopefully your $\Box$ operator is diagonalizable with your choice of boundary conditions.

One nice aspect of your redefinition of the $\log$ is that if $\Box = A + \delta A$, where $A$ might be easy to diagonalize but $A+\delta A$ not so much and where $[A, \delta A] \neq 0$, then your formula provides a nice rewriting of the logarithm that does not involve messy details of Baker-Campbell-Hausdorff.

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  • $\begingroup$ Thanks for your answer. Do you know any other reference for all of this besides wikipedia? I need to see this in more detail. It would be great to see the Lie theory perspective for this. Which lie group does the operator box benlong to? $\endgroup$
    – user74106
    Commented Dec 9, 2015 at 11:16
  • $\begingroup$ I would have thought it's only natural to associate $\Box$ to a Lie group when the manifold is a symmetric space or a coset space. In this case, $\Box$ should be related to the Casimirs of the associated simple Lie groups. (I love Wikipedia, but I would guess a vanilla book on linear algebra might be useful. For symmetric spaces, there is an ancient tome by Helgason.) $\endgroup$ Commented Dec 9, 2015 at 15:40

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