ncg

week9.tex (18001B)

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 \title{University of Vienna\\ Faculty of Physics\\ \vspace{1.25cm}
 Notes on\\ Noncommutative Geometry and Particle Phyiscs}
 \author{Milutin Popovic \\ Supervisor: Dr. Lisa
 Glaser}
 \date{Week 9: 28.05 - 4.06}
 
 \begin{document}
 
     \maketitle
     \tableofcontents
     \newpage
 
 \section{Heat Kernel Expansion}
 \subsection{The Heat Kernel}
 The heat kernel $K(t; x, y; D)$ is the fundamental solution of the heat
 equation. It depends on the operator $D$ of Laplacian type.
 \begin{align}
     (\partial _t + D_x)K(t;x, y;D) =0
 \end{align}
 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}
 we have 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
 guage field, the heat kernel reads then
 \begin{align}
     K(t;x,y;D) = \langle x|e^{-tD}|y\rangle.
 \end{align}
 We can expand it it in terms of $D_0$ and we still have the
 singularity from the equation \ref{eq:standard} as $t\rightarrow 0$ thus the
 expansion gives
 \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 $b_k(x,y)$ are regular in $y \rightarrow x$. They are called the heat
 kernel coefficients.
 
 \subsection{Example}
 Now let us consider a propagator $D^{-1}(x,y)$ defined through the heat kernel
 in an integral representation
 \begin{align}
     D^{-1} (x,y) = \int_0^\infty dt K(t;x,y;D).
 \end{align}
 We can integrate the expression formally if we assume the heat kernel vanishes
 for $t\rightarrow \infty$ we 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}
 it solves the 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 integral approximation of the propagator we conclude
 that the singularities of $D^{-1}$ coincide with the singularities of the heat
 kernel coefficients.
 We consider now a generating functional in terns of $\det(D)$ which is called
 the one-loop effective action (quantum fields theory)
 \begin{align}
     W = \frac{1}{2}\ln(\det D)
 \end{align}
 we can relate $W$ with the heat kernel. 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 depent on
 $\lambda$, because of this we can ignore it. Further more we use $\ln(\det D) =
 \text{Tr}(\ln D)$ and therefor we can write for $W$
 \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 is now 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 are just ignored. The divergences at $t\rightarrow 0$
 are cutoff at $t=\Lambda^{-2}$, thus we write
 \begin{align}
     W_\Lambda = -\frac{1}{2} \int_{\Lambda^{-2}}^\infty dt \frac{K(t, D)}{t}.
 \end{align}
 We can calculate $W_\Lambda$ at 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 $b_2(x,x)$ with $k\leq n$. Now we compute the limit
 $\Lambda \rightarrow \infty$
 \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}
 here $\Gamma$ is the gamma function.
 \subsection{Differential Geometry and Operators of Laplace Type}
 Let $M$ be a $n$ dimensional compact Riemannian manifold with $\partial M = 0$.
 Then consider a vector bundle $V$ over $M$ (i.e. there is a vector space to
 each point on $M$), so we can define smooth functions. We want to look at
 arbitrary differential operators $D$ of Laplace type on $V$, they have the general
 from
 \begin{align}
     D = -(g^{\mu\nu} \partial_\mu\partial_\nu + a^\sigma\partial_\sigma +b)
 \end{align}
 where $a^\sigma, b$ are matrix valued functions on $M$ and $g^{\mu\nu}$ is the
 inverse metric on $M$. There is a unique connection on $V$ and a unique
 endomorphism (matrix valued function) $E$ on $V$, then we can rewrite $D$ in
 terms of $E$ and covariant derivatives
 \begin{align}
     D = -(g^{\mu\nu} \nabla_\mu \nabla_\nu +E)
 \end{align}
 Where the covariant derivative consists of $\nabla = \nabla^{[R]} +\omega$ the
 standard Riemannian covariant derivative $\nabla^{[R]}$ and a "gauge" bundle
 $\omega$ (fluctuations). WE can write $E$ and $\omega$ in terms of geometrical
 identities
 \begin{align}
     \omega_\delta &= \frac{1}{2}g_{\nu\delta}(a^\nu
     +g^{\mu\sigma}\Gamma^\nu_{\mu\sigma}I_V)\\
     E &= b - g^{\nu\mu}(\partial_\mu \omega_\nu + \omega_\nu \omega_\mu -
     \omega_\sigma \Gamma^\sigma_{\nu\mu})
 \end{align}
 where $I_V$ is the identity in $V$ and the Christoffel symbol
 \begin{align}
     \Gamma^\sigma_{\mu\nu} = g^{\sigma\varrho} \frac{1}{2} (\partial_\mu
     g_{\nu\varrho} + \partial_\nu g_{\mu\varrho} - \partial_\varrho g_{\mu\nu})
 \end{align}
 Furthermore we remind ourselves of the Riemmanian curvature tensor, Ricci
 Tensor and the Scalar curavture.
 \begin{align}
     R^\mu_{\nu\varrho\sigma} &= \partial_\sigma \Gamma^{\mu}_{\nu\varrho}
     -\partial_\varrho \Gamma^\mu_{\nu\sigma}
     \Gamma^{\lambda}_{\nu\varrho}\Gamma^{\mu}_{\lambda\sigma}
     \Gamma^{\lambda}_{\nu\sigma}\Gamma^{\mu}_{\lambda\varrho}\\
     R_{\mu\nu} &:= R^{\sigma}_{\mu\nu\sigma}\\
     R &:= R^\mu_{\ \mu}
 \end{align}
 
 The we let $\{e_1, \dots, e_n\}$ be the local orthonormal frame of
 $TM$(tangent bundle $M$), which will be noted with flat indices $i,j,k,l
 \in\{1,\dots, n\}$, we use $e^k_\mu, e^\nu_j$ to transform between flat indices
 and curved indices $\mu, \nu, \varrho$.
 \begin{align}
     e^\mu_j e^\nu_k g_{\mu\nu} &= \delta_{jk}\\
     e^\mu_j e^\nu_k \delta^{jk} &= g^{\mu\nu} \\
     e^j_\mu e^\mu_k  &= \delta^j_k
 \end{align}
 
 The Riemannian part of the covariant derivative contains the standard
 Levi-Civita connection, so that for a $v_\nu$ we write
 \begin{align}
     \nabla_\mu^{[R]} v_\nu = \partial_\mu v_\nu -
     \Gamma^{\varrho}_{\mu\nu}v_\varrho.
 \end{align}
 The extended covariant derivative reads then
 \begin{align}
     \nabla_\mu v^j = \partial_\mu v^j + \sigma^{jk}_\mu v_k.
 \end{align}
 the condition $\nabla_\mu e^k_\nu = 0$ gives us the general connection
 \begin{align}
     \sigma^{kl}_\mu = e^\nu_l\Gamma^{\varrho}_{\mu\nu}e^k_\varrho - e^\nu_l
     \partial_\mu e^k_\nu
 \end{align}
 The we may define the field strength $\Omega_{\mu\nu}$ of the connection $\omega$
 \begin{align}
     \Omega_{\mu\nu} = \partial_\mu \omega_\nu -\partial_\nu \omega_\mu
     +\omega_\mu \omega_\nu -\omega_\nu\omega_\mu.
 \end{align}
 If we apply the covariant derivative on $\Omega$ we get
 \begin{align}
     \nabla_\varrho\Omega_{\mu\nu} = \partial_\varrho \Omega_{\mu\nu} -
     \Gamma^{\sigma}_{\varrho \mu} \Omega_{\sigma\mu} + [\omega_\varrho,
     \Omega_{\mu\nu}]
 \end{align}
 
 \subsection{Spectral Functions}
 Manifolds without $M$ boundary condition for the operator $e^{-tD}$ for $t>0$ is a
 trace class operator on $L^2(V)$, this means that for any smooth function $f$
 on $M$ we can define
 \begin{align}
     K(t,f,D) = \text{Tr}_{L^2}(fe^{-tD})
 \end{align}
 and we can rewrite
 \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 terms of the Heat kernel $K(t;x,y;D)$ in the regular limit $y\rightarrow y$.
 We can write the Heat Kernel in terms of the spectrum of $D$. Say
 $\{\phi_\lambda\}$ is a ONB of eigenfunctions of $D$ corresponding to the
 eigenvalue $\lambda$
 \begin{align}
     K(t;x,y;D) = \sum_\lambda \phi^\dagger_\lambda(x)
     \phi_\lambda(y)e^{-t\lambda}.
 \end{align}
 We have an asymtotic expansion at $t \rightarrow 0$  for the trace
 \begin{align}
     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}
 \subsection{General Formulae}
 We consider a compact Riemmanian Manifold $M$ without boundary condition, a
 vector bundle $V$ over $M$ to define functions which carry discrete (spin or
 gauge) indices. An Laplace style operator $D$ over $V$ and smooth function $f$
 on $M$. There is an asymtotic expansion where the heat kernel coefficients
 \begin{enumerate}
     \item with odd index $k=2j+1$ vanish
         $a_{2j+1}(f,D) = 0$
     \item with even index are locally computable in terms of geometric
         invariants
 \end{enumerate}
 \begin{align}
     a_k(f,D) &= \text{Tr}_V\left(\int_M d^n x\sqrt{g}(f(x)a_k(x;D)\right) =\\
     &=\sum_I \text{Tr}_V\left(\int_M d^nx \sqrt{g}(fu^I \mathcal{A}^I_k(D))\right)
 \end{align}
 here $\mathcal{K}^I_k$ are all possible independent invariants of dimension
 $k$, constructed from $E, \Omega, R_{\mu\nu\varrho\sigma}$ and their
 derivatives, $u^I$ are some constants.
 
 If $E$ has dimension two, then the derivative has dimension one. So 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 $D = D_1 \otimes 1 + 1 \otimes D_2$ we can
 separate everything, i.e.
 \begin{align}
     e^{-tD} &= e^{-tD_1} \otimes e^{-tD_2}\\
     f(x_1, x_2) &= f_1(x_1)f_2(x_2)\\
     a_k(x;D) &= \sum_{p+q=k} a_p(x_1; D_1)a_q(x_2;D_2)
 \end{align}
 Say the spectrum of $D_1$ is known, $l^2, l\in \mathbb{Z}$. We obtain the heat
 kernel asymmetries with the Poisson Summation formula
 \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}} = \\
     &\simeq \sqrt{\frac{\pi}{t}} + \mathcal{O}(e^{-1/t}).
 \end{align}
 Note that the exponentially small terms have no effect on the heat kernel
 coefficients and that the only nonzero coefficient is $a_0(1, D_1) =
 \sqrt{\pi}$. Therefore we can write
 \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}
 
 On the other had all geometric invariants associated with $D$ are in the $D_2$
 part. Thus all invariants are independent of $x_1$, so we can choose for $M_1$.
 Say $M_1 = S^1$ with $x\in (0, 2\pi)$ and $D_1=-\partial_{x_1}^2$ we may
 rewrite the heat kernel coefficients in
 \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))=\\
     &= 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}
 
 \subsection{Heat Kernel Coefficients}
 To calculate the heat kernel coefficients we need the following variational
 equations
 \begin{align}
     &\frac{d}{d\varepsilon}|_{\varepsilon=0}a_k(1, e^{-2\varepsilon f}D) =
     (n-k) a_k(f, D),\label{eq:var1}\\
     &\frac{d}{d\varepsilon}|_{\varepsilon=0}a_k(1, D-\varepsilon F) =
     a_{k-2}(F,D),\label{eq:var2}\\
     &\frac{d}{d\varepsilon}|_{\varepsilon=0}a_k(e^{-2\varepsilon f}F,
     e^{-2\varepsilon f}D) =
     0\label{eq:var3}.
 \end{align}
 To prove the equation \ref{eq:var1} we differentiate
 \begin{align}
     \frac{d}{d\varepsilon}|_{\varepsilon=0} \text{Tr}(\exp(-e^{-2\varepsilon
     f}tD) = \text{Tr}(2ftDe^{-tD}) = -2t\frac{d}{dt}\text{Tr}(fe^{-tD}))
 \end{align}
 then we expand both sides in $t$ and get \ref{eq:var1}. Equation \ref{eq:var2}
 is derived similarly. For equation \ref{eq:var3} we consider the following
 operator
 \begin{align}
     D(\varepsilon,\delta) = e^{-2\varepsilon f}(D-\delta F)
 \end{align}
 for $k=n$ we use equation \ref{eq:var1} and we get
 \begin{align}
     \frac{d}{d\varepsilon}|_{\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}|_{\delta=0}\frac{d}{d\varepsilon}|_{\varepsilon=0}a_n(1,
     D(\varepsilon,\delta)) =
     \frac{d}{d\varepsilon}|_{\varepsilon=0}\frac{d}{d\delta}|_{\delta=0}a_n(1,
     D(\varepsilon,\delta)) =\\
     &=a_{n-2} ( e^{-2\varepsilon f}F, e^{-2\varepsilon f}D)
 \end{align}
 which proves equation \ref{eq:var3}. With this we calculate the constants $u^I$
 and we can write the first three heat kernel coefficients as
 \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 RE + \alpha_5 E^2
     \alpha_6 R_{,kk} + \\
     &+\alpha_7 R^2 + \alpha_8 R_{ij}R_{ij} + \alpha_9
     R_{ijkl}R_{ijkl} +\alpha_{10} \Omega_{ij}\Omega{ij})).
 \end{align}
 The constants $\alpha_I$ do not depend on the dimension $n$ of the Manifold and
 we can compute them with our variational identities.
 
 The first coefficient $\alpha_0$ can be seen from the heat kernel expanion of
 the Laplacian on $S^1$ (above), $\alpha_0 = 1$. For $\alpha_1$ we use
 \ref{eq:var2}, for $k = 2$
 \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}
 thus we conclude that $\alpha_1 = 6$. Now we take $k=4$
 \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 FE + \alpha_2 FR),
 \end{align}
 thus $\alpha_4 = 60\alpha_2$ and $\alpha_5 = 180$.
 
 Furthermore we apply \ref{eq:var3} to $n=4$
 \begin{align}
     \frac{d}{d\varepsilon}|_{\varepsilon=0} a_2(e^{-2\varepsilon f}F,
     e^{-2\varepsilon f}D) = 0.
 \end{align}
 By 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 $M_1, M_2$, then we can decompose the heat
 kernel coefficients for $k=4$
 \begin{align}
     a_4(1,-\Delta_1-\Delta_2) =& a_4(1, -\Delta_1) a_0(1, -\Delta_2)
     +a_2(1,-\Delta_1) a_2(1,-\Delta_2) \\&+ 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 +\\
     &+(2\alpha_3 + 10\alpha_6)f_{,iijj}+\\
     &+(2\alpha_4 -2\alpha_6 - 20\alpha_7 -2\alpha_8)f_{,ii}R\\
     &+(-8\alpha_8 -8\alpha_6)f_{,ij}R_{ij}))
 \end{align}
 we obtain $\alpha_3 = 60$, $\alpha_6=12$, $\alpha_8 = -2$ and $\alpha_9 = 2$
 
 For $\alpha_{10}$ we use the Gauss-Bonnet theorem to get $\alpha_{10}=30$,
 which is left out because it is a lengthy computation.
 
 Summarizing we get for the 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}))\\
 \end{align}
 
 
 
 
 
 
 
 
 
 \end{document}