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I wish to compute the most general tensor Gaussian integral of the form

$$\int [\mathcal{D}A] \exp(-A^{\mu_1\mu_2\cdots \mu_s}M_{\mu_1\mu_2\cdots\mu_s\nu_1\nu_2\cdots\nu_s}A^{\nu_1\nu_2\cdots \nu_s})$$

My guess is that the answer should be $$\det(M)^{-1/2}$$ where $\det{M}$ is defined here but I am unable to prove it. Can anybody provide a rigorous way to do such an integral?

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    $\begingroup$ In the correct basis this should become a product of single variable gaussian integrals. $\endgroup$
    – Guy
    Commented Apr 21, 2022 at 19:18
  • $\begingroup$ Is $A$ just a tensor, or is it a tensor field? In other words, is this just a matrix (tensor) integral or is it actually a path integral? In the former case, I would say that neither of the tags quantum-field-theory or path-integral is appropriate and that the question would be more suitable for math.SE. In the latter case, can you rewrite the action in a more transparent way? $\endgroup$ Commented Apr 21, 2022 at 20:58
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    $\begingroup$ Hint: raise the manifold dimension to the power of $s$. $\endgroup$
    – J.G.
    Commented Apr 21, 2022 at 21:24
  • $\begingroup$ Take s =2 first. $\endgroup$ Commented Apr 22, 2022 at 20:36

1 Answer 1

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Just rename indices, $\mu_1\mu_2\cdots \mu_s\to i$ (that is, $11\cdots 11\to 1, \ 11\cdots12\to2$, etc, $nn\cdots nn\to n^s$. Then your integral is $$ \int e^{-A_i M_{ij} A_j}=\underset{ij}{\det}(M_{ij})^{-1/2} $$

This is also $$ \underset{ij}{\det}(M_{ij})\epsilon_{j_1\cdots j_N}=\epsilon^{i_1\cdots i_N}M_{i_1j_1}M_{i_2,j_2}\cdots M_{i_N,j_N} $$ which you can "unfold" into your vector indices, $i_k\to \mu_{1,k}\mu_{2,k}\cdots\mu_{s,k}$, if you so desire.

This is how computers deal with higher-dimensional arrays, after all.

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  • $\begingroup$ Are you trying to say $M_{ij}$ is now an $n^s \times n^s$ matrix and the gaussian integral is now the determinant of this matrix? And what do you mean by unfold? $\endgroup$ Commented Apr 24, 2022 at 6:53
  • $\begingroup$ Also, is your answer same as my guess or not? $\endgroup$ Commented Apr 24, 2022 at 7:04
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    $\begingroup$ @user44690 1) Yes. 2) unfold = undo the renaming. 3) It is not clear to me what "your guess" is. The formulas in your linked answer do not make much sense to me. My naive interpretation of what they mean does not seem to be equivalent to what I wrote, though. $\endgroup$ Commented Apr 24, 2022 at 16:57

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