I am looking for a way to prove that $$ \det (M \cdot N) = \det(M)\det(N) \tag{0}$$ Where $M$ and $N$ are matrices with continuous indices, so that $\det$ is a functional determinant. A way to show that $(0)$ is wrong would also be welcomed.

This question is about the following formula, $$ \int\text{d}\vec{x} \exp(- \sum_{ij}x^i A_{ij}x^j) = \left (\det A_{ij}\right )^{-1/2}\left (2\pi\right )^{D/2}. \tag{1} $$ Now, we would like this identity to be compatible with, $$ \int\text{d}\vec{x} \exp(- \sum_{ijk}x^i A_{ik}B_{kj}x^j) = \left (\det A\cdot B\right )^{-1/2}\left (2\pi\right )^{D/2} = \left (\det A\right )^{-1/2}\left (\det B\right )^{-1/2}\left (2\pi\right )^{D/2}.\tag{2} $$ Any idea how to prove this? I am interested, eventually, in the generalisation of this formula to path integrals, namely, given the path integral
$$ \int\mathcal{D}\phi \exp\left[- \int\text{d}x\text{d}y \phi(x)M(x,y)\phi(y)\right] =C \left (\det M\right )^{-1/2}, \tag{3} $$ where now $\det M$ is a functional determinant, i ask the question whether it makes sense to write the generalised formula, $$\begin{align} \int\mathcal{D}\phi \exp\left[- \int\text{d}x\text{d}y \text{d}z\phi(x)M(x,y)N(y,z)\phi(z)\right] =& \left (\det M\cdot N\right )^{-1/2}\cr =& \left (\det M\right )^{-1/2} \left (\det N\right )^{-1/2}.\end{align} \tag{4} $$

[UPDATE]: I might have an answer now: let us just consider, $$\det M\cdot N = \prod_i \lambda_i[M\cdot N],\tag{5}$$ where $\lambda_i[M\cdot N]$ are the the eigenvalues of the matrix $M\cdot N$. This formula is valid even for continuous matrices, such as the laplacian operator $\partial^2 \delta(x-y)$. If the commutator $[M,N] = 0$, then the two matrices can be diagonalised in the same basis, and $\lambda_i[M\cdot N] = \lambda_i[M]\lambda_i[N]$, with no sum over $i$. Then formula (4) can be proven at least in the simple case in which the commutator vanishes. A trivial example of this is for $M = A$ and $N = A^{-1}$, for any invertible matrix $A$, which leads to $\det A\cdot A^{-1}=1$. Also, in case $M\cdot M^T = f(x) \delta(x-y)$, this would imply that $$\det M\cdot M^T = (\det M)^2 = \det f(x) \delta(x-y) = \prod_x f(x)\tag{6}$$ and so on. These seem trivial cases, but since we are talking of functional determinants they constitute a powerful computational tool. How much do you agree with this attempt of a solution? It is not very formal, but i don't see where it could go wrong.

  • $\begingroup$ Can't you assume you bring $A_{ij}$ to its diagonal form by a suitable change of $x$, which will not affect the integration because the measure is invariant? $\endgroup$ – ZeroTheHero Mar 10 '17 at 14:51
  • $\begingroup$ @ZeroTheHero But what you say only works for $A$ diagonalizable, and I don't think that's true in general for this formula. $\endgroup$ – Aaron Mar 10 '17 at 15:36
  • $\begingroup$ @Aaron yes I assumed $A$ diagonalizable since $\sum_{ij} x^iA_{ij}x^j$ looks like a quadratic form. $\endgroup$ – ZeroTheHero Mar 10 '17 at 15:40
  • $\begingroup$ @ZeroTheHero Ah you are completely correct; $A$ can WLOG be assumed to be symmetric because it's in a quadratic form. $\endgroup$ – Aaron Mar 10 '17 at 15:44
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    $\begingroup$ @MrFermiMr can't you then just take $M=A\cdot B$ and proceed as per your first formula, which would give you just $det(M)$ which you can then expand as $det(A\cdot B)=det(A)\det(B)$? (BTW... great question) $\endgroup$ – ZeroTheHero Mar 10 '17 at 16:15

The statement seems to be wrong even for an infinite number of discrete indices.

Consider for example the vector space of square integrable functions on the positive integers, i.e. sequences $\{f_1,f_2,\cdots\}$ s.t. $\sum_{i>0} |f_i|^2 < \infty$, and consider the shift operator $S:f \mapsto Sf$, where

$Sf = \{ f_2,f_3,\cdots \} \ . $

Consider furthermore the operator $S^\dagger: f \mapsto S^\dagger f$ with

$S^\dagger f = \{ 0, f_1,f_2, \cdots \} \ . $


$SS^\dagger f = \{f_1,f_2,\cdots \} = f \ \ \ \text{ but } \ \ \ S^\dagger S f = \{ 0, f_2,f_3,\dots \} \ . $

That is, $SS^\dagger$ has all eigenvalues $1$, while $S^\dagger S$ has one eigenvalue $0$ and all other eigenvalues $1$. Hence

$ \det( SS^\dagger) = 1 \ \ \ \ \ \text{while} \ \ \ \ \det (S^\dagger S ) = 0 \ .$

Now if it were true that $\det(M N) = \det(M) \det(N)$, than we would have proven that $1 = 0$.

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The following comments seem relevant to OP's problem:

  1. For a matrix $A\in{\rm Mat}_{n\times n}(\mathbb{C})$, define the symmetrized matrix $$A_+~:=~ \frac{A+A^T}{2}.\tag{A}$$

  2. Then the Gaussian integral reads $$ \int_{\mathbb{R}^n} \! d^n x ~e^{-\frac{1}{2} x^T A x} ~=~ \sqrt{\frac{(2\pi)^n}{\det A_+}}\tag{B}$$ if the matrix ${\rm Re}A_+$ is positive definite, cf. e.g. this math.SE post.

  3. Similarly, $$ \int_{\mathbb{R}^n} \! d^n x ~e^{-\frac{1}{2} x^T AB x} ~=~ \sqrt{\frac{(2\pi)^n}{\det (AB)_+}}\tag{C}$$ if the matrix ${\rm Re}(AB)_+$ is positive definite, cf. OP's eq. (2).

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  • $\begingroup$ Sure, i can then rewrite (C) as $$ \sqrt{1/(2\pi)^n} \sqrt{(2\pi)^n/\det(A_+)} \sqrt{(2\pi)^n/\det( B_+)} \qquad (C1)$$ because of the property of the determinant. My question is if there is an alternative way to prove this identity, without using the determinant property. This is to generalise (C1) to functional determinants. $\endgroup$ – MrFermiMr Mar 10 '17 at 16:29
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    $\begingroup$ Forget the infinite-dimensional case for starters. The rewriting (C1) is in general not valid already in finite dimensions. In particular, the relation $(AB)_+=A_+B_+$ is in general not true. (Hint: Consider e.g. the antisymmetric case $B=A=-A^T$. Then $A_+=0=B_+$ while $AB$ is symmetric.) $\endgroup$ – Qmechanic Mar 10 '17 at 16:37
  • $\begingroup$ @Qumechanic can you elaborate? You're saying that $\det (A\cdot B) ) = (\det A)(\det B)$? I am pretty sure that this is true for discrete matrices. $\endgroup$ – MrFermiMr Mar 10 '17 at 16:44
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    $\begingroup$ @MrFermiMr the problem is that your identity (2) is wrong in general even for finite dimensional matrices. $\endgroup$ – Lorenz Mayer Oct 22 '18 at 10:58

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