Asymptotic expansion of heat operator $e^{-Delta{t}}$ and $e^{-mathcal{D}t}$ of Dirac operator












2












$begingroup$


For a closed Riemannian manifold $M$ of $n$-dimension, we consider the Laplace-Beltrami operator $Delta$.



It is known that we have an asymptotic expansion for the trace of heat operator $e^{-Delta{t}}$ as follows
$$
mathrm{tr}(e^{-Delta{t}})=sum_{lambda}e^{-lambda{t}}overset{tdownarrow0}{sim}t^{-frac{n}{2}}sum_{n} alpha_{n}t^{n},
$$
where $lambda$ runs over the set of spectrum of Laplacian $Delta$.



My question is that




Denote by $mathcal{D}$ the Dirac operator whose square coincides with the Laplacian, $i.e.$ $mathcal{D}^{2}=Delta$.
Then the sum of positive eigenvalues of an operator $e^{-mathcal{D}t}$ $$sum_{lambdainmathrm{Sp}(Delta)}e^{-sqrt{lambda}{t}}$$ has an asymptotic expansion around $t=0$?
If it exists, then is it possible to induce a relation between coefficients?




I know that the proof for the case of heat operator follows from the construction of heat kernel. But I wonder that the same construction can be applied to the Dirac operator.



Thank you for your time and effort.






differential-geometry pde asymptotics spectral-theory differential-operators







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    For a closed Riemannian manifold $M$ of $n$-dimension, we consider the Laplace-Beltrami operator $Delta$.



    It is known that we have an asymptotic expansion for the trace of heat operator $e^{-Delta{t}}$ as follows
    $$
    mathrm{tr}(e^{-Delta{t}})=sum_{lambda}e^{-lambda{t}}overset{tdownarrow0}{sim}t^{-frac{n}{2}}sum_{n} alpha_{n}t^{n},
    $$
    where $lambda$ runs over the set of spectrum of Laplacian $Delta$.



    My question is that




    Denote by $mathcal{D}$ the Dirac operator whose square coincides with the Laplacian, $i.e.$ $mathcal{D}^{2}=Delta$.
    Then the sum of positive eigenvalues of an operator $e^{-mathcal{D}t}$ $$sum_{lambdainmathrm{Sp}(Delta)}e^{-sqrt{lambda}{t}}$$ has an asymptotic expansion around $t=0$?
    If it exists, then is it possible to induce a relation between coefficients?




    I know that the proof for the case of heat operator follows from the construction of heat kernel. But I wonder that the same construction can be applied to the Dirac operator.



    Thank you for your time and effort.






    differential-geometry pde asymptotics spectral-theory differential-operators







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      For a closed Riemannian manifold $M$ of $n$-dimension, we consider the Laplace-Beltrami operator $Delta$.



      It is known that we have an asymptotic expansion for the trace of heat operator $e^{-Delta{t}}$ as follows
      $$
      mathrm{tr}(e^{-Delta{t}})=sum_{lambda}e^{-lambda{t}}overset{tdownarrow0}{sim}t^{-frac{n}{2}}sum_{n} alpha_{n}t^{n},
      $$
      where $lambda$ runs over the set of spectrum of Laplacian $Delta$.



      My question is that




      Denote by $mathcal{D}$ the Dirac operator whose square coincides with the Laplacian, $i.e.$ $mathcal{D}^{2}=Delta$.
      Then the sum of positive eigenvalues of an operator $e^{-mathcal{D}t}$ $$sum_{lambdainmathrm{Sp}(Delta)}e^{-sqrt{lambda}{t}}$$ has an asymptotic expansion around $t=0$?
      If it exists, then is it possible to induce a relation between coefficients?




      I know that the proof for the case of heat operator follows from the construction of heat kernel. But I wonder that the same construction can be applied to the Dirac operator.



      Thank you for your time and effort.






      differential-geometry pde asymptotics spectral-theory differential-operators







      share|cite|improve this question









      $endgroup$




      For a closed Riemannian manifold $M$ of $n$-dimension, we consider the Laplace-Beltrami operator $Delta$.



      It is known that we have an asymptotic expansion for the trace of heat operator $e^{-Delta{t}}$ as follows
      $$
      mathrm{tr}(e^{-Delta{t}})=sum_{lambda}e^{-lambda{t}}overset{tdownarrow0}{sim}t^{-frac{n}{2}}sum_{n} alpha_{n}t^{n},
      $$
      where $lambda$ runs over the set of spectrum of Laplacian $Delta$.



      My question is that




      Denote by $mathcal{D}$ the Dirac operator whose square coincides with the Laplacian, $i.e.$ $mathcal{D}^{2}=Delta$.
      Then the sum of positive eigenvalues of an operator $e^{-mathcal{D}t}$ $$sum_{lambdainmathrm{Sp}(Delta)}e^{-sqrt{lambda}{t}}$$ has an asymptotic expansion around $t=0$?
      If it exists, then is it possible to induce a relation between coefficients?




      I know that the proof for the case of heat operator follows from the construction of heat kernel. But I wonder that the same construction can be applied to the Dirac operator.



      Thank you for your time and effort.






      differential-geometry pde asymptotics spectral-theory differential-operators




      differential-geometry pde asymptotics spectral-theory differential-operators






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 30 '18 at 11:54









      Junhyeong KimJunhyeong Kim

      558




      558






















          1 Answer
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          0












          $begingroup$

          We define the spectral zeta function of the Laplacian $Delta$ by
          $$
          zeta_{Delta}(s) = sum_{lambdainmathrm{Sp}(Delta)}lambda^{-s}.
          $$
          It is known that $zeta_{Delta}(s)$ converges absolutely over $mathrm{Re}(s)>frac{n}{2}$, and has an analytic continuation to a meromorphic function over all complex plane.
          It can be expressed by
          $$zeta_{Delta}(s) = frac{1}{Gamma(s)}int_{0}^{infty}t^{s-1}sum_{lambdainmathrm{Sp}(Delta)}e^{-lambda{t}}dt,$$
          where $Gamma(s):=int_{0}^{infty}t^{s-1}e^{-t}dt$ is the Gamma function.



          From the identities
          $$
          Gamma(s)lambda^{-s}=int_{0}^{infty}t^{s-1}e^{-lambda{t}}dt
          $$
          and
          $$
          Gamma(2s)lambda^{-s}=int_{0}^{infty}t^{2s-1}e^{-sqrt{lambda}t}dt,
          $$
          we deduce the following relation
          $$
          zeta_{Delta}(s) = frac{1}{Gamma(s)}int_{0}^{infty}t^{s-1}sum_{lambdainmathrm{Sp}(Delta)}e^{-lambda{t}}dt \
          =frac{1}{Gamma(2s)}int_{0}^{infty}t^{2s-1}sum_{lambdainmathrm{Sp}(Delta)}e^{-sqrt{lambda}t}dt.
          $$
          Since the spectral zeta function $zeta_{Delta}(s)$ is meromorphic over a complex plane, $i.e.$ is analytic except for discrete singularities, the series
          $$
          sum_{lambdainmathrm{Sp}(Delta)}e^{-sqrt{lambda}t}
          $$
          has an asymptotic expansion.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            If the spectral zeta function is meromorphic over a complex plane and $zeta_Delta(s) Gamma(2s)$ satisfies the conditions of some tauberian theorem then the series has an asymptotic expansion given by the poles of $zeta_Delta(s) Gamma(2s)$. (try with $-1 + sum_n e^{-n^2 x}$ whose Mellin transform $Gamma(s) zeta(2s)$ has only one pole)
            $endgroup$
            – reuns
            Dec 8 '18 at 20:24













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          1 Answer
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          1 Answer
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          0












          $begingroup$

          We define the spectral zeta function of the Laplacian $Delta$ by
          $$
          zeta_{Delta}(s) = sum_{lambdainmathrm{Sp}(Delta)}lambda^{-s}.
          $$
          It is known that $zeta_{Delta}(s)$ converges absolutely over $mathrm{Re}(s)>frac{n}{2}$, and has an analytic continuation to a meromorphic function over all complex plane.
          It can be expressed by
          $$zeta_{Delta}(s) = frac{1}{Gamma(s)}int_{0}^{infty}t^{s-1}sum_{lambdainmathrm{Sp}(Delta)}e^{-lambda{t}}dt,$$
          where $Gamma(s):=int_{0}^{infty}t^{s-1}e^{-t}dt$ is the Gamma function.



          From the identities
          $$
          Gamma(s)lambda^{-s}=int_{0}^{infty}t^{s-1}e^{-lambda{t}}dt
          $$
          and
          $$
          Gamma(2s)lambda^{-s}=int_{0}^{infty}t^{2s-1}e^{-sqrt{lambda}t}dt,
          $$
          we deduce the following relation
          $$
          zeta_{Delta}(s) = frac{1}{Gamma(s)}int_{0}^{infty}t^{s-1}sum_{lambdainmathrm{Sp}(Delta)}e^{-lambda{t}}dt \
          =frac{1}{Gamma(2s)}int_{0}^{infty}t^{2s-1}sum_{lambdainmathrm{Sp}(Delta)}e^{-sqrt{lambda}t}dt.
          $$
          Since the spectral zeta function $zeta_{Delta}(s)$ is meromorphic over a complex plane, $i.e.$ is analytic except for discrete singularities, the series
          $$
          sum_{lambdainmathrm{Sp}(Delta)}e^{-sqrt{lambda}t}
          $$
          has an asymptotic expansion.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            If the spectral zeta function is meromorphic over a complex plane and $zeta_Delta(s) Gamma(2s)$ satisfies the conditions of some tauberian theorem then the series has an asymptotic expansion given by the poles of $zeta_Delta(s) Gamma(2s)$. (try with $-1 + sum_n e^{-n^2 x}$ whose Mellin transform $Gamma(s) zeta(2s)$ has only one pole)
            $endgroup$
            – reuns
            Dec 8 '18 at 20:24


















          0












          $begingroup$

          We define the spectral zeta function of the Laplacian $Delta$ by
          $$
          zeta_{Delta}(s) = sum_{lambdainmathrm{Sp}(Delta)}lambda^{-s}.
          $$
          It is known that $zeta_{Delta}(s)$ converges absolutely over $mathrm{Re}(s)>frac{n}{2}$, and has an analytic continuation to a meromorphic function over all complex plane.
          It can be expressed by
          $$zeta_{Delta}(s) = frac{1}{Gamma(s)}int_{0}^{infty}t^{s-1}sum_{lambdainmathrm{Sp}(Delta)}e^{-lambda{t}}dt,$$
          where $Gamma(s):=int_{0}^{infty}t^{s-1}e^{-t}dt$ is the Gamma function.



          From the identities
          $$
          Gamma(s)lambda^{-s}=int_{0}^{infty}t^{s-1}e^{-lambda{t}}dt
          $$
          and
          $$
          Gamma(2s)lambda^{-s}=int_{0}^{infty}t^{2s-1}e^{-sqrt{lambda}t}dt,
          $$
          we deduce the following relation
          $$
          zeta_{Delta}(s) = frac{1}{Gamma(s)}int_{0}^{infty}t^{s-1}sum_{lambdainmathrm{Sp}(Delta)}e^{-lambda{t}}dt \
          =frac{1}{Gamma(2s)}int_{0}^{infty}t^{2s-1}sum_{lambdainmathrm{Sp}(Delta)}e^{-sqrt{lambda}t}dt.
          $$
          Since the spectral zeta function $zeta_{Delta}(s)$ is meromorphic over a complex plane, $i.e.$ is analytic except for discrete singularities, the series
          $$
          sum_{lambdainmathrm{Sp}(Delta)}e^{-sqrt{lambda}t}
          $$
          has an asymptotic expansion.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            If the spectral zeta function is meromorphic over a complex plane and $zeta_Delta(s) Gamma(2s)$ satisfies the conditions of some tauberian theorem then the series has an asymptotic expansion given by the poles of $zeta_Delta(s) Gamma(2s)$. (try with $-1 + sum_n e^{-n^2 x}$ whose Mellin transform $Gamma(s) zeta(2s)$ has only one pole)
            $endgroup$
            – reuns
            Dec 8 '18 at 20:24
















          0












          0








          0





          $begingroup$

          We define the spectral zeta function of the Laplacian $Delta$ by
          $$
          zeta_{Delta}(s) = sum_{lambdainmathrm{Sp}(Delta)}lambda^{-s}.
          $$
          It is known that $zeta_{Delta}(s)$ converges absolutely over $mathrm{Re}(s)>frac{n}{2}$, and has an analytic continuation to a meromorphic function over all complex plane.
          It can be expressed by
          $$zeta_{Delta}(s) = frac{1}{Gamma(s)}int_{0}^{infty}t^{s-1}sum_{lambdainmathrm{Sp}(Delta)}e^{-lambda{t}}dt,$$
          where $Gamma(s):=int_{0}^{infty}t^{s-1}e^{-t}dt$ is the Gamma function.



          From the identities
          $$
          Gamma(s)lambda^{-s}=int_{0}^{infty}t^{s-1}e^{-lambda{t}}dt
          $$
          and
          $$
          Gamma(2s)lambda^{-s}=int_{0}^{infty}t^{2s-1}e^{-sqrt{lambda}t}dt,
          $$
          we deduce the following relation
          $$
          zeta_{Delta}(s) = frac{1}{Gamma(s)}int_{0}^{infty}t^{s-1}sum_{lambdainmathrm{Sp}(Delta)}e^{-lambda{t}}dt \
          =frac{1}{Gamma(2s)}int_{0}^{infty}t^{2s-1}sum_{lambdainmathrm{Sp}(Delta)}e^{-sqrt{lambda}t}dt.
          $$
          Since the spectral zeta function $zeta_{Delta}(s)$ is meromorphic over a complex plane, $i.e.$ is analytic except for discrete singularities, the series
          $$
          sum_{lambdainmathrm{Sp}(Delta)}e^{-sqrt{lambda}t}
          $$
          has an asymptotic expansion.






          share|cite|improve this answer









          $endgroup$



          We define the spectral zeta function of the Laplacian $Delta$ by
          $$
          zeta_{Delta}(s) = sum_{lambdainmathrm{Sp}(Delta)}lambda^{-s}.
          $$
          It is known that $zeta_{Delta}(s)$ converges absolutely over $mathrm{Re}(s)>frac{n}{2}$, and has an analytic continuation to a meromorphic function over all complex plane.
          It can be expressed by
          $$zeta_{Delta}(s) = frac{1}{Gamma(s)}int_{0}^{infty}t^{s-1}sum_{lambdainmathrm{Sp}(Delta)}e^{-lambda{t}}dt,$$
          where $Gamma(s):=int_{0}^{infty}t^{s-1}e^{-t}dt$ is the Gamma function.



          From the identities
          $$
          Gamma(s)lambda^{-s}=int_{0}^{infty}t^{s-1}e^{-lambda{t}}dt
          $$
          and
          $$
          Gamma(2s)lambda^{-s}=int_{0}^{infty}t^{2s-1}e^{-sqrt{lambda}t}dt,
          $$
          we deduce the following relation
          $$
          zeta_{Delta}(s) = frac{1}{Gamma(s)}int_{0}^{infty}t^{s-1}sum_{lambdainmathrm{Sp}(Delta)}e^{-lambda{t}}dt \
          =frac{1}{Gamma(2s)}int_{0}^{infty}t^{2s-1}sum_{lambdainmathrm{Sp}(Delta)}e^{-sqrt{lambda}t}dt.
          $$
          Since the spectral zeta function $zeta_{Delta}(s)$ is meromorphic over a complex plane, $i.e.$ is analytic except for discrete singularities, the series
          $$
          sum_{lambdainmathrm{Sp}(Delta)}e^{-sqrt{lambda}t}
          $$
          has an asymptotic expansion.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 7 '18 at 13:03









          Junhyeong KimJunhyeong Kim

          558




          558












          • $begingroup$
            If the spectral zeta function is meromorphic over a complex plane and $zeta_Delta(s) Gamma(2s)$ satisfies the conditions of some tauberian theorem then the series has an asymptotic expansion given by the poles of $zeta_Delta(s) Gamma(2s)$. (try with $-1 + sum_n e^{-n^2 x}$ whose Mellin transform $Gamma(s) zeta(2s)$ has only one pole)
            $endgroup$
            – reuns
            Dec 8 '18 at 20:24




















          • $begingroup$
            If the spectral zeta function is meromorphic over a complex plane and $zeta_Delta(s) Gamma(2s)$ satisfies the conditions of some tauberian theorem then the series has an asymptotic expansion given by the poles of $zeta_Delta(s) Gamma(2s)$. (try with $-1 + sum_n e^{-n^2 x}$ whose Mellin transform $Gamma(s) zeta(2s)$ has only one pole)
            $endgroup$
            – reuns
            Dec 8 '18 at 20:24


















          $begingroup$
          If the spectral zeta function is meromorphic over a complex plane and $zeta_Delta(s) Gamma(2s)$ satisfies the conditions of some tauberian theorem then the series has an asymptotic expansion given by the poles of $zeta_Delta(s) Gamma(2s)$. (try with $-1 + sum_n e^{-n^2 x}$ whose Mellin transform $Gamma(s) zeta(2s)$ has only one pole)
          $endgroup$
          – reuns
          Dec 8 '18 at 20:24






          $begingroup$
          If the spectral zeta function is meromorphic over a complex plane and $zeta_Delta(s) Gamma(2s)$ satisfies the conditions of some tauberian theorem then the series has an asymptotic expansion given by the poles of $zeta_Delta(s) Gamma(2s)$. (try with $-1 + sum_n e^{-n^2 x}$ whose Mellin transform $Gamma(s) zeta(2s)$ has only one pole)
          $endgroup$
          – reuns
          Dec 8 '18 at 20:24




















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