ncg

4_heatkernel.tex (13630B)

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 \subsection{Heat Kernel Expansion\label{sec:4}}
 \subsubsection{The Heat Kernel}
 The heat kernel $K(t; x, y; D)$ is the fundamental solution to the heat
 equation
 \begin{align}
     (\partial _t + D_x)K(t;x, y;D) =0,
 \end{align}
 which depends on the operator $D$ of Laplacian type.
 
 For a flat manifold $M = \mathbb{R}^n$ and $D = D_0 := -\Delta_\mu\Delta^\mu +m^2$ the
 Laplacian with a mass term and the initial condition
 \begin{align}
     K(0;x,y;D) = \delta(x,y),
 \end{align}
 takes the form of the standard fundamental solution
 \begin{align}\label{eq:standard}
     K(t;x,y;D_0) = (4\pi t)^{-n/2}\exp\left(-\frac{(x-y)^2}{4t}-tm^2\right).
 \end{align}
 
 Let us consider now a more general operator $D$ with a potential term or a
 gauge field, the heat kernel then reads
 \begin{align}
     K(t;x,y;D) = \langle x|e^{-tD}|y\rangle.
 \end{align}
 We can expand the heat kernel in $t$, still having a
 singularity from the equation \eqref{eq:standard} as $t \rightarrow 0$, on
 obtains
 \begin{align}
     K(t;x,y;D) = K(t;x,y;D_0)\left(1 + tb_2(x,y) + t^2b_4(x,y) + \dots
     \right),
 \end{align}
 where $b_k(x,y)$ become regular as $y \rightarrow x$. These coefficients are called the heat
 kernel coefficients.
 %%----------------------- KANN WEGGELASSEN WERDEN
 %\newline
 %\textbf{KANN WEGELASSEN WERDEN BIS ZUM NÄCHSTEN KAPITEL}
 %Let's turn our attention to a propagator $D^{-1}(x,y)$ defined through the
 %heat kernel, with an integral representation
 %\begin{align}
 %    D^{-1} (x,y) = \int_0^\infty dt K(t;x,y;D).
 %\end{align}
 %If we assume the heat kernel vanishes for $t\rightarrow \infty$, we can
 %integrate formally to get
 %\begin{align}
 %    D^{-1}(x,y) \simeq
 %    2(4\pi)^{-n/2}\sum_{j=0}\left(\frac{|x-y|}{2m}\right)^{-\frac{n}{2}+j+1}
 %    K_{-\frac{n}{2}+j+1}(|x-y|m)b_{2j}(x,y),
 %\end{align}
 %where $b_0 = 1$ and $K_\nu (z)$ is the Bessel function
 %\begin{align}
 %    K_\nu(z) = \frac{1}{\pi} \int_0^\pi \cos(\nu\tau-z\sin(\tau))d\tau.
 %\end{align}
 %The Bessel function solves the following differential equation
 %\begin{align}
 %    z^2 \frac{d^2K}{dz^2} + z \frac{dK}{dz} + (z^2 - \nu^2)=0.
 %\end{align}
 %By looking at an integral approximation for the propagator we conclude that
 %the singularities of $D^{-1}$ coincide with the singularities of the heat
 %kernel coefficients. Thus we can say, that a generating functional in terms of
 %$\det(D)$ is called the one-loop effective action (quantum field theory)
 %\begin{align}
 %    W = \frac{1}{2}\ln(\det D).
 %\end{align}
 %We have a direct relation with one-loop effective action $W$ and the
 %heat kernel. Furthermore notice that for each eigenvalue $\lambda >0$ of $D$
 %we can write the identity.
 %\begin{align}
 %    \ln \lambda  = -\int_0^\infty \frac{e^{-t\lambda}}{t}dt
 %\end{align}
 %This expression is correct up to an infinite constant which does not depend
 %on the eigenvalue $\lambda$, thus we can ignore it. By substituting
 %$\ln(\det D) = \text{Tr}(\ln D)$ we can rewrite the one-loop effective action
 %$W$ into
 %\begin{align}
 %    W = -\frac{1}{2} \int_0^\infty dt \frac{K(t, D)}{t},
 %\end{align}
 %where
 %\begin{align}
 %    K(t, D) = \text{Tr}(e^{-tD}) = \int d^n x \sqrt{g}K(t;x,x;D).
 %\end{align}
 %The problem now is that the integral of $W$ is divergent at both limits. Yet
 %the divergences at $t\rightarrow \infty$ are caused by $\lambda \leq 0$ of $D$
 %(infrared divergences) and can be ignored. The divergences at $t\rightarrow 0$
 %are cutoff at $t=\Lambda^{-2}$, simply written as
 %\begin{align}
 %    W_\Lambda = -\frac{1}{2} \int_{\Lambda^{-2}}^\infty dt \frac{K(t, D)}{t}.
 %\end{align}
 %We can calculate $W_\Lambda$ up to an order of $\lambda ^0$
 %\begin{align}
 %    W_\Lambda &= -(4\pi)^{-n/2} \int d^n x\sqrt{g}\bigg(
 %    \sum_{2(j+l)<n}\Lambda^{n-2j-2l}b_{2j}(x,x) \frac{(-m^2)^l l!}{n-2j-2l} +\\
 %    &+ \sum_{2(j+l) =n }\ln(\Lambda) (-m^2)^l l! b_{2j}(x,x)
 %    \mathcal{O}(\lambda^0) \bigg)
 %\end{align}
 %There is an divergence at $b_2(x,x)$ for $k\leq n$. Computing the limit
 %$\Lambda \rightarrow \infty$ we get
 %\begin{align}
 %    -\frac{1}{2}(4\pi)^{n/2}m^n\int d^n x\sqrt{g} \sum_{2j>n}
 %    \frac{b_{2j}(x,x)}{m^{2j}}\Gamma(2j-n),
 %\end{align}
 %where $\Gamma$ stands for the gamma function.
 %%----------------------- KANN WEGGELASSEN WERDEN
 
 
 \subsubsection{Spectral Functions}
 Manifolds $M$ with a disappearing boundary condition for the operator
 $e^{-tD}$ for $t>0$, i.e.\ a trace class operator on $L^2(V)$. Meaning for any
 smooth function $f$ on $M$ the Heat kernel can be defined as
 \begin{align}
     K(t,f,D) := \text{Tr}_{L^2}(fe^{-tD}).
 \end{align}
 Alternately an integral representation is
 \begin{align}
     K(t, f, D) = \int_M d^n x \sqrt{g} \text{Tr}_V(K(t;x,x;D)f(x)),
 \end{align}
 in the regular limit $y \rightarrow y$. The Heat Kernel can be written in terms
 of the spectrum of $D$. Hence for an orthonormal basis $\{\phi_\lambda\}$ of
 eigenfunctions for $D$, which correspond to the eigenvalue $\lambda$ one
 can rewrite the heat kernel into
 \begin{align}
     K(t;x,y;D) = \sum_\lambda \phi^\dagger_\lambda(x)
     \phi_\lambda(y)e^{-t\lambda}.
 \end{align}
 An asymptotic expansion as $t \rightarrow 0$ for the trace is then
 \begin{align}
     \text{Tr}_{L^2}(fe^{-tD}) \simeq \sum_{k\geq 0}t^{(k-n)/2}a_k(f,D),
 \end{align}
 where
 \begin{align}
     a_k(f,D) = (4\pi)^{-n/2} \int_M d^4x \sqrt{g} b_k(x,x) f(x).
 \end{align}
 \subsubsection{General Formulae}
 Let us summarize what we have obtained in the last chapter. We considered a
 compact Riemannian manifold $M$ without boundary condition, a vector bundle
 $V$ over $M$ to define functions which carry discrete (spin or gauge)
 indices, an operator $D$ of Laplace type over $V$ and smooth function $f$ on
 $M$.
 
 There is an asymptotic expansion where the heat kernel coefficients with an
 odd index $k=2j+1$ vanish $a_{2j+1}(f,D) = 0$. On the other hand coefficients
 with an even index are locally computable in terms of geometric invariants
 \begin{align}
     a_k(f,D) &= \text{Tr}_V\left(\int_M d^n x\sqrt{g}(f(x)a_k(x;D)\right)
     =\nonumber\\
     &=\sum_I \text{Tr}_V\left(\int_M d^nx \sqrt{g}(fu^I
     \mathcal{A}^I_k(D))\right).
 \end{align}
 The notation $\mathcal{A}^I_k$ corresponds to all possible independent invariants of
 dimension $k$ and $u^I$ are constants. The invariants are constructed from
 $E, \Omega, R_{\mu\nu\varrho\sigma}$ and their derivatives. If $E$ has
 dimension two, then the derivative has dimension one. In this way if $k=2$ there are
 only two independent invariants, $E$ and $R$. This corresponds to the
 statement $a_{2j+1}=0$.
 
 If we consider $M = M_1 \times M_2$ with coordinates $x_1$ and $x_2$ and a
 decomposed Laplace style operator $D = D_1 \otimes 1 + 1 \otimes D_2$ the
 functions acting on operators and on coordinates are separable linearly by the
 following
 \begin{align}
     e^{-tD} &= e^{-tD_1} \otimes e^{-tD_2},\\
     f(x_1, x_2) &= f_1(x_1)f_2(x_2),
 \end{align}
 thus the heat kernel coefficients are separated by
 \begin{align}
     a_k(x;D) &= \sum_{p+q=k} a_p(x_1; D_1)a_q(x_2;D_2).
 \end{align}
 Lets say the eigenvalues of $D_1$ are $l^2, l\in \mathbb{Z}$, we
 can obtain the heat kernel asymmetries with the Poisson summation formula
 giving us an approximation in the order of $e^{-1/t}$
 \begin{align}
     K(t, D_1) &= \sum_{l\in\mathbb{Z}} e^{-tl^2} = \sqrt{\frac{\pi}{t}}
     \sum_{l\in\mathbb{Z}} e^{-\frac{\pi^2l^2}{t}} = \nonumber \\
     &\simeq \sqrt{\frac{\pi}{t}} + \mathcal{O}(e^{-1/t}).
 \end{align}
 The exponentially small terms have no effect on the heat kernel
 coefficients and the only nonzero coefficient is $a_0(1, D_1) =
 \sqrt{\pi}$, therefore the heat coefficients can be written as
 \begin{align}
     a_k(f(x^2), D) = \sqrt{\pi}\int_{M_2}
     d^{n-1}x\sqrt{g}\sum_I\text{Tr}_V\left(f(x^2)u^I_{(n-1)}
     \mathcal{A}^I_n(D_2)\right).
 \end{align}
 
 Since all of the geometric invariants  associated with $D$ are in the $D_2$
 part, they are independent of $x_1$. This allows us to to choose for
 $M_1$.
 
 For $M_1 = S^1$ with $x\in (0, 2\pi)$ and $D_1=-\partial_{x_1}^2$
 we can rewrite the heat kernel coefficients into
 \begin{align}
     a_k(f(x_2), D) &= \int_{S^1\times M_2}d^nx \sqrt{g} \sum_I
     \text{Tr}_V(f(x_2) u_{(n)}^I \mathcal{A}^I_k(D_2))= \nonumber\\
     &= 2\pi \int_{M_2} d^nx\sqrt{g} \sum_I\text{Tr}_V(f(x_2) u_{(n)}^I
     \mathcal{A}^I_k(D_2)).
 \end{align}
 Computing the two equations above we see that
 \begin{align}
     u_{(n)}^I = \sqrt{4\pi} u^I_{(n+1)}.
 \end{align}
 
 \subsubsection{Heat Kernel Coefficients}
 To calculate the heat kernel coefficients consider the following variational
 equations
 \begin{align}
     &\frac{d}{d\varepsilon}\bigg|_{\varepsilon=0}a_k(1, e^{-2\varepsilon f}D) =
     (n-k) a_k(f, D),\label{eq:var1}\\
     &\frac{d}{d\varepsilon}\bigg|_{\varepsilon=0}a_k(1, D-\varepsilon F) =
     a_{k-2}(F,D),\label{eq:var2}\\
     &\frac{d}{d\varepsilon}\bigg|_{\varepsilon=0}a_k(e^{-2\varepsilon f}F,
     e^{-2\varepsilon f}D) =
     0\label{eq:var3}.
 \end{align}
 Let us explain the equations above. To get the first equation \eqref{eq:var1}
 we simply differentiate
 \begin{align}
     \frac{d}{d\varepsilon}\bigg|_{\varepsilon=0} \text{Tr}(\exp(-e^{-2\varepsilon
     f}tD) = \text{Tr}(2ftDe^{-tD}) = -2t\frac{d}{dt}\text{Tr}(fe^{-tD})),
 \end{align}
 additionally expand both sides in $t$ and get equation \eqref{eq:var1}. Equation
 \eqref{eq:var2} is derived similarly.
 
 For equation \eqref{eq:var3} look at the following operator
 \begin{align}
     D(\varepsilon,\delta) = e^{-2\varepsilon f}(D-\delta F),
 \end{align}
 for $k=n$ we take equation \eqref{eq:var1} and we obtain
 \begin{align}
     \frac{d}{d\varepsilon}\bigg|_{\varepsilon=0}a_n(1,D(\varepsilon,\delta))
     =0.
 \end{align}
 Then we take the variation in terms of $\delta$, evaluated at $\delta =0$ and
 swap the differentiation, allowed by theorem of Schwarz
 \begin{align}
     0 &=
     \frac{d}{d\delta}\bigg|_{\delta=0}\frac{d}{d\varepsilon}\bigg|_{\varepsilon=0}a_n(1,
     D(\varepsilon,\delta)) =\nonumber\\
       &=\frac{d}{d\varepsilon}\bigg|_{\varepsilon=0}\frac{d}{d\delta}\bigg|_{\delta=0}a_n(1,
     D(\varepsilon,\delta)) =\nonumber\\
       &=a_{n-2} ( e^{-2\varepsilon f}F, e^{-2\varepsilon f}D),
 \end{align}
 which gives us equation \eqref{eq:var3}.
 
 Now that the ground basis is established, we can calculate the constants
 $u^I$, and by that the first three heat kernel coefficients read
 \begin{align}
     a_0(f, D) &= (4\pi)^{-n/2}\int_Md^n x\sqrt{g} \text{Tr}_V(a_0 f),\\
     a_2(f, D) &= (4\pi)^{-n/2}\frac{1}{6}\int_Md^n
     x\sqrt{g}\text{Tr}_V)(f\alpha _1 E+\alpha _2 R),\\
     a_4(f, D) &= (4\pi)^{-n/2}\frac{1}{360}\int_Md^n
     x\sqrt{g}\text{Tr}_V(f(\alpha_3 E_{,kk} + \alpha_4\ R\ E + \alpha_5 E^2
     \alpha_6 R_{,kk} + \nonumber\\
     &+\alpha_7 R^2 + \alpha_8 R_{ij}R_{ij} + \alpha_9
     R_{ijkl}R_{ijkl} +\alpha_{10} \Omega_{ij}\Omega{ij})),
 \end{align}
 where the comma subscript $,$ denotes the derivative and constants $\alpha_I$
 do not depend on the dimension of the Manifold and we can compute them with
 our variational identities.
 
 The first coefficient $\alpha_0$ can be read from the heat kernel expansion of
 the Laplacian on $S^1$ (above), $\alpha_0 = 1$. For $\alpha_1$ we use
 \eqref{eq:var2}, the coefficient $k = 2$ is
 \begin{align}
     \frac{1}{6} \int_M d^n x\sqrt{g} \text{Tr}_V(\alpha_1F) = \int_M d^n
     x\sqrt{g} \text{Tr}_V(F),
 \end{align}
 which means $\alpha_1 = 6$. Looking at the coefficient $k=4$ we have
 \begin{align}
     \frac{1}{360}\int_Md^n x \sqrt{g}\text{Tr}_V(\alpha_4\ F\ R + 2\alpha_5\ F\ E)
     = \frac{1}{6} \int_Md^n x\sqrt{g}\text{Tr}_V(\alpha_1\ F\ E + \alpha_2\ F\ R),
 \end{align}
 thus $\alpha_4 = 60\alpha_2$ and $\alpha_5 = 180$.
 
 By applying  \eqref{eq:var3} to $n=4$ we get
 \begin{align}
     \frac{d}{d\varepsilon}|_{\varepsilon=0} a_2(e^{-2\varepsilon f}F,
     e^{-2\varepsilon f}D) = 0.
 \end{align}
 Collecting the terms with $\text{Tr}_V(\int_Md^nx\sqrt{g}(Ff_{,jj}))$ we
 obtain $\alpha_1 = 6\alpha_2$, that is $\alpha_2 = 1$, so $\alpha_4 = 60$.
 
 Now we let $M=M_1\times M_2$ and split $D = -\Delta_1 -\Delta_2$, where
 $\Delta_{1/2}$ are Laplacians for $M_1, M_2$. This allows us to decompose the heat
 kernel coefficient for $k=4$ into
 \begin{align}
     a_4(1,-\Delta_1-\Delta_2) =& a_4(1, -\Delta_1) a_0(1,
     -\Delta_2)\nonumber+ \\
                                &+a_2(1,-\Delta_1) a_2(1,-\Delta_2)\nonumber \\
                                &+ a_0(1,-\Delta_1)a_4(1,-\Delta_2),
 \end{align}
 with $E=0$ and $\Omega =0$ and by calculating the terms with $R_1R_2$  (scalar
 curvature of $M_{1/2}$) we obtain $\frac{2}{360}\alpha_7 =
 (\frac{\alpha_2}{6})^2$, thus $\alpha_7 = 5$.
 
 For $n=6$ we get
 \begin{align}
     0 &= \text{Tr}_V(\int_Md^nx\sqrt{g}
     (F(-2\alpha_3-10\alpha_4+4\alpha_5)f_{,kk}E +\nonumber\\
     &+(2\alpha_3 + 10\alpha_6)f_{,iijj}+\nonumber\\
     &+(2\alpha_4 -2\alpha_6 - 20\alpha_7 -2\alpha_8)f_{,ii}R\nonumber\\
     &+(-8\alpha_8 -8\alpha_6)f_{,ij}R_{ij}))
 \end{align}
 we obtain $\alpha_3 = 60$, $\alpha_6=12$, $\alpha_8 = -2$ and $\alpha_9 = 2$
 
 To get $\alpha_{10}$ we use the Gauss-Bonnet theorem, ultimately giving us
 $\alpha_{10}=30$. We leave out this lengthy calculation and refer to
 \cite{heatkernel} for further reading.
 
 Let us summarize our calculations which ultimately lead us to the following heat kernel
 coefficients
 \begin{align}
     \alpha_0(f, D) &= (4\pi)^{-n/2}\int_M d^n x \sqrt{g} \text{Tr}_V(f),\\
     \alpha_2(f, D) &= (4\pi)^{-n/2}\frac{1}{6}\int_M d^n x \sqrt{g}
     \text{Tr}_V(f(6E+R)),\\
     \alpha_4(f, D) &= (4\pi)^{-n/2}\frac{1}{360}\int_M d^n x \sqrt{g}
     \text{Tr}_V(f(60E_{,kk}+60RE+ 180E^2 +\\
     &+12R_{,kk} + 5R^2 - 2 R_{ij}R_{ij}
     2R_{ijkl}R_{ijkl} +30\Omega_{ij}\Omega_{ij})).\label{eq: a_4}
 \end{align}