Question: Given an arbitrary number of real matrices of the form $ A_i= \biggl(\begin{matrix} C_i+E_i & B_i \\ B_i^T & D_i-F_i \end{matrix} \biggr) $, where $B_i$ is an arbitrary $n\times n$ real matrix, $C_i$ and $D_i$ are $n\times n$ real anti-symmetric matrices, $E_i$ and $F_i$ are $n\times n$ real symmetric and positive semidefinite matrices, how to prove the following $$ \det \biggl(I_{2n} + \prod_i e^{A_i}\biggl)\ge 0 \,? $$

Background: This is a stronger version of an earlier MO question How to prove this determinant is positive? which was solved by GH from MO and Terry Tao. Their proof addressed the case of $E_i=F_i=0$, where $e^{A_i}$ belongs to the split orthogonal group.

The paper arXiv:1601.01994v2 in fact contains a rigorous proof of the above statement using tools familiar to physicists, e.g. Majorana fermion and reflection positivity (cf Eq.(10)). It would be nice to see an alternate mathematical proof (perhaps of a similar nature to the split orthogonal group proof by GH from MO and Terry Tao).

Addendum: The paper arXiv:1601.01994v2 contains further results on complex matrices, which I do not know how to formulate in simple math language yet.

  • $\begingroup$ I find it interesting such a many-body problem hinges on such a basic result $\endgroup$ Feb 1 '16 at 12:35
  • $\begingroup$ Maybe one can start with a warm up question with $B_i=C_i=D_i=0$ ? Then the statement becomes $det(I_n +\prod_i e^{E_i}) * det(I_n +\prod_i e^{-F_i}) \ge 0$, where $E_i$ and $F_i$ are real symmetric and positive semidefinite matrices. $\endgroup$
    – Lei Wang
    Feb 1 '16 at 14:34
  • $\begingroup$ to the warm up : $det(I_n +\prod_i e^{-F_i}) \ge 0$ because $\Vert\prod_i e^{-F_i}\Vert \le 1$ and $F_i$ real . Then also $det(I_n +\prod_i e^{E_i}) = det(\prod_i e^{E_i}) det(I_n +(\prod_i e^{E_i})^{-1}) \ge 0$ . $\endgroup$
    – jjcale
    Feb 3 '16 at 19:49
  • $\begingroup$ The $e^{A_i}$ increase the quadratic form that the split orthogonal group leaves invariant . So maybe this is the generalization of Part I : The split orthogonal group is replaced by the semi group that increases the quadratic form. $\endgroup$
    – jjcale
    Feb 4 '16 at 18:50
  • $\begingroup$ @jjcale Thanks! Could you elaborate a bit more about "increase the quadratic from" ? $\endgroup$
    – Lei Wang
    Feb 5 '16 at 7:57

Let $q(x,y) = x^H J y$ for $x,y \in \mathbb{C}^{2n}$ where $J = diag(I_n,-I_n)$ and let $S = \{A \in M_{2n}(\mathbb{R}) : q(Ax, Ax) \ge q(x,x) $ $\forall x \in \mathbb{C}^{2n}\}$. Obviously $S$ is a semi group . Furthermore the $e^{t A_i}$ are in $S$ since $$\frac{d}{dt} q(e^{t A_i} x,e^{t A_i} x) = 2 (e^{t A_i} x)^H diag (E_i,F_i) (e^{t A_i} x) \ge 0$$ for all $x \in \mathbb{C}^{2n}$.

Now let $T : [0,1] \rightarrow S$ analytic where $det(I_{2n}+T(0)) > 0$ and where $T(0)$ has no degenerate eigenvalues.

Let $E_{\lambda}(t)$ be the generalized eigenspace of $T(t)$ to the eigenvalue $\lambda$ . For $G \subset \mathbb{C}$ define $$E_G(t) = \bigoplus_{\lambda \in G}{E_{\lambda}(t)}$$ .

Let $t_0 \in [0,1]$ such that $det(I_{2n}+T(t_0)) = 0$ .

Now we want to show that $dim\,E_{(-1,\infty)}(t)$ can only change by an even number near $t_0$ . Therefore $det(I_{2n}+T(t))$ can't change the sign.

Lemma 1 : Let $U \in S$, $x \in \mathbb{C}^{2n}$ with $0 = q(x,x) = q(Ux,Ux)$ . Then $q(x,y) = q(Ux,Uy)$ for all $y \in \mathbb{C}^{2n}$ .

Proof : We have $q(ax+y,ax+y) \le q(a Ux + Uy,a Ux + Uy)$ for all $a \in \mathbb{C}$ and therefore $0 \le 2 Re\,a (q(Uy,Ux) - q(y,x)) + q(Uy,Uy) - q(y,y)$ . But the right hand side can be made negative for appropriate a if $q(Uy,Ux) \neq q(y,x)$ .

Lemma 2 : Let $p$ a polynomial and $z \in \mathbb{C}$ . If $p(n) z^n$ is constant for all large enough $n \in \mathbb{N}$ then $p$ is constant and $z = 1$ or $p = 0$ or $z = 0$.

Proof left to the reader.

Lemma 3 : Let $U \in S$, $x$ a generalized eigenvector of $U$ to the eigenvalue $\lambda$, $y$ a generalized eigenvector of $U$ to the eigenvalue $\mu$ and $q(U^l x,U^l x) = 0$ for all $l \in \mathbb{N}_0$. Then holds $\lambda \bar{\mu} = 1$ or $q(x,y) = 0$ .

Proof : By Lemma 1 we have $q(y,x) = q(U^l y, U^l x)$ for all $l \in \mathbb{N}$ . But $q(U^l y, U^l x)$ has the form $p(l) (\lambda \bar{\mu})^l$ for a polynomial $p$ for all large enough $l \in \mathbb{N}$ . From Lemma 2 then follows Lemma 3 .

Lemma 4 : Let $U \in S$, $x$ a generalized eigenvector of $U$ to the eigenvalue $-1$ and $q(x,U^k x) = 0$ for all $k \in \mathbb{N}_0$ . Then $q(U^k x,U^l x) = 0$ for all $k,l \in \mathbb{N}_0$ .

Proof : Let $x_k = (I_{2n} + U)^k x$ and $m$ minimal such that $q(x_j,x_k) = 0$ for all $j,k \geq m$ . First we want to show that $q(x_j,x_k) = 0$ for $j \geq m$ and $k \geq 0$. If this is not the case then let $j \geq m$ and $k$ be maximal such that $q(x_j,x_k) \neq 0$ . Then $k > 0$ and by Lemma 1 $q(x_j,x_{k-1}) = q(U^l x_j,U^l x_{k-1}) = q(x_j,x_{k-1}) - l q(x_j,x_k)$ for all $l \in \mathbb{N}$ . Contradiction ! Now we get for $m > 1$ $q(x_{m-2},x_{m-2}) \leq q(U^l x_{m-2},U^l x_{m-2}) = l^2 q(x_{m-1},x_{m-1}) + O(l)$ and $q(x_{m-2},x_{m-2}) \geq q(U^{-l} x_{m-2},U^{-l} x_{m-2}) = l^2 q(x_{m-1},x_{m-1}) + O(l)$ . Contradiction to $q(x_{m-1},x_{m-1}) \neq 0$ ! Since $m = 1$ is impossible since $q(x_0,x_k) = 0$ for all $k \geq 0$ we are done .

Lemma 5 : The restriction of $q$ to $E_{-1}(t_0)$ is non degenerate .

Proof : Let $x \in E_{-1}(t_0)$ and $U = T(t_0)$ . We want to show that there exists $y \in E_{-1}(t_0)$ such that $q(x,y) \neq 0$ . If there exists $n \in \mathbb{N}_0$ such that $q(x,U^n x) \neq 0$ we are done. Otherwise ist follows from Lemma 4 and Lemma 3 that x is orthogonal w.r.t. q to all other generalized eigenspaces . But since q is non degenerate there exists $y \in E_{-1}(t_0)$ such that $q(x,y) \neq 0$ .

Lemma 6 : T(t) has degenerated eigenvalues only at isolated points.

Proof : The discriminant of T(t) is analytic in t and nonzero at t = 0 .

Now we can choose $\epsilon > 0$ and $r$ with $0 < r < 1$ such that for $\vert t-t_0\vert < \epsilon$ holds :

i) $det(I_{2n}+T(t)) \neq 0$ for $t \neq t_0$ ,

ii) $E_{\{z: \vert z+1 \vert \leq r\}}(t_0) = E_{-1}(t_0)$ ,

iii) $\sigma (T(t)) \cap \{z: \vert z+1 \vert = r\} = \emptyset$ ,

iv) the signature of the restriction of $q$ to $V(t)$ is constant where $V(t) = E_{\{z: \vert z+1 \vert \leq r\}}(t)$ ,

v) T(t) has no degenerate eigenvalues for $t \neq t_0$ .

Let $D = \{z: \vert z+1 \vert \leq r\}$ . For each eigenvalue $\lambda \in D$ of $T(t)$ with $\vert \lambda \vert = 1$ we can write $E_{\lambda}(t) = E_{\lambda}^+(t) \oplus E_{\lambda}^-(t)$ such that the restriction of $q$ to $E_{\lambda}^+(t)$ is positive semidefinite and the restriction of $q$ to $E_{\lambda}^-(t)$ is negative definite and such that $E_{\bar{\lambda}}^+(t) = \overline{E_{\lambda}^+(t)}$ and $E_{\bar{\lambda}}^-(t) = \overline{E_{\lambda}^-(t)}$ .

Now we can write $V(t) = V^+(t) \oplus V^-(t)$ where $$V^+(t) = E_{D \cap \{z : \vert z \vert > 1\}}(t) \oplus \bigoplus_{\lambda \in D , \vert \lambda \vert = 1} E_{\lambda}^+(t)$$ and $$V^-(t) = E_{D \cap \{z : \vert z \vert < 1\}}(t) \oplus \bigoplus_{\lambda \in D , \vert \lambda \vert = 1} E_{\lambda}^-(t)$$ .

Now we want to show that for $\vert t-t_0\vert < \epsilon$ and $t \neq t_0$ the restriction of $q$ to $V^+(t)$ is positive semidefinite and the restriction to $V^-(t)$ is negative semidefinite :

For $x \in V^+(t)$ we get $$q(x,x) \geq \lim_{m \rightarrow \infty} \frac{1}{m} \sum_{l=1}^m q(U^{-l} x,U^{-l} x) = \sum_{\lambda \in D, \vert \lambda \vert = 1} q(x_{\lambda}^+,x_{\lambda}^+) \geq 0$$ where $x_{\lambda}^+$ is the component of x in $E_{\lambda}^+(t)$ . For $x \in V^-(t)$ we get $$q(x,x) \leq \lim_{m \rightarrow \infty} \frac{1}{m} \sum_{l=1}^m q(U^l x,U^l x) = \sum_{\lambda \in D, \vert \lambda \vert = 1} q(x_{\lambda}^-,x_{\lambda}^-) \leq 0$$ where $x_{\lambda}^-$ is the component of x in $E_{\lambda}^-(t)$ .

Let $n_+$ the number of positive eigenvalues of the restriction of q to $V(t)$ and $n_-$ the number of negative eigenvalues. We have shown that $n_+ \geq dim\, V^+(t)$ and $n_- \geq dim\, V^-(t)$ and therefore $n_+ = dim\, V^+(t)$ . Since $dim\, V^+(t) - dim\, E_{(-1,-1-r)}$ is even $dim\, E_{(-1,-1-r)}$ can only change by an even number. And $dim\, E_{[-1-r,-\infty)}$ can only change if a pair of complex conjugate eigenvalues gets real or vice versa and therefore also only by an even number.

So we have shown that $dim\,E_{(-1,\infty)}(t)$ can only differ by an even number on different points in $\vert t-t_0\vert < \epsilon, t\neq t_0$ .

To finish the proof, choose $W$ such that $e^{t W} \in S$ for $t \in [0,1]$ and such that $e^W$ fullfills the requirements on $T(0)$ . Then choose $$T(t) = e^{(1-t) W} \prod_i e^{t A_i}$$ .

  • $\begingroup$ Thanks a lot! I still need some time to fully digest your proof. Anyway, let me ask the following question first: according to your proof, do you think the statement is already the strongest one can make? In other words, do you see any chance to further generalize it ? Thanks! $\endgroup$
    – Lei Wang
    Feb 26 '16 at 10:10
  • 2
    $\begingroup$ What one needs is $A_i^T J + J A_i \ge 0$ but I haven't checked whether this is a generalization. $\endgroup$
    – jjcale
    Mar 2 '16 at 18:01
  • $\begingroup$ Oh I see. This definitely generalized the split orthogonal group condition $A_i^TJ + J A_i=0$. $\endgroup$
    – Lei Wang
    Mar 4 '16 at 0:53

Even though this paper [1] is in the strongly correlated electrons section of the cond-mat (and it's about Quantum Monte Carlo) we can learn a bit about Hermitian matrices. In particular, I argue that reflection-positivity is a natural matrix notion if you're reading a paper like this.

In Supplement I we get general form of the density matrix. Though I am not sure about the size of this matrix:

$$ \rho_P = \mathrm{Tr} \prod_{k=1}^M e^{-\tau H_0} e^{-\eta H_I(\eta_k)} = \det \left( I + \prod_{k=1}^M e^{-\tau h^0} e^{-\eta h^I (\eta_k)} \right)$$

This strikes me as ambiguous, since this trace is also a determinant, and in fact $\rho_P$ is a matrix. I suspect this has to do with the "Fermions" which are a way of arranging Hilbert space into certain tensor products.

Eq (5) and (6) say either of $H_0$ or $H_I$ could take the "reflection symmetric" form:

$$ H = \gamma^T \left( \begin{array}{cc} A & iB \\ -iB^T & A^* \end{array}\right) \gamma $$

Although $A,B$ are $2N \times 2N$ matrices, it's unclear about the shape of $H$. We have $\gamma = (\gamma_i^{(1)}, \gamma_i^{(2)}) $ which lives in a $2\times N$ dimensional space and I am still not sure how to write the tensor product.

Then in supplement 3A we make the choices:

$$ A = \frac{1}{2} \left( \begin{array}{cc} C & - iF \\ i F^T & D \end{array}\right) \hspace{0.25in} B = \frac{1}{2} \left( \begin{array}{cc} B_1 & 0\\ 0 & B_2 \end{array}\right)$$

And I have still have not addressed the change of basis between Dirac and Majorana fermions.

If we set $\eta = 0$ and $\tau = -1$, theorem 1 says if $H_0$ is ``majorana reflection positive" then:

$$ \rho_P = \mathrm{Tr} \prod_{k=1}^M e^{-\tau H_0} = \det \left( I + \prod_{k=1}^M e^{-\tau h^0} \right)$$

However, it's ambiguous what to put for $H_0$ or $h^0$, possibly eq (10).

Certainly, your result is much stronger that $\rho_P \geq 0$ is a positive semi-definite matrix.


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