Given an orthogonal matrix $O$ with dimensions $4n \times 4n$ and $\det O = -1$, how to prove that $\det[O_{11} - O_{22} + i (O_{12} + O_{21})] = 0$?

Here $O$ is a block matrix $[[O_{11}, O_{12}], [O_{21}, O_{22}]]$, and all blocks have equal size.

An equivalent statement is the following: if $\det O = -1$, then $\det(\Omega - O \Omega O^T)=0$, with $\Omega = [[0, I],[-I, 0]]$. The reason why I believe this is correct is due to the physical meaning of it (it's about presence of bound states in a certain system), and numerical checks. I am not sure why odd $n$ doesn't work, but it makes sense for physics reasons.

Example of a matrix for which it works is below:

$$\left(\begin{array}{cccc} 0.44090815& -0.71206204& -0.44576549& 0.31600755 \\\ 0.12767731& 0.35584235& 0.19884606& 0.90417641 \\\ 0.88038152& 0.229104 & 0.30519084& -0.28159952 \\\ 0.11927654& 0.56022784& -0.81768693& -0.05749749 \end{array}\right)$$

Edit(Will Jagy):In particular, this fails for $(4n - 2) \times (4n - 2),$ as in the $ 2 \times 2$ $$ O \; \; = \; \; \left( \begin{array}{cc} \cos t & \sin t\\\ \sin t & - \cos t \end{array} \right) , $$ with the relevant $ 1 \times 1$ matrix of complex numbers $$ \left( \cos t - ( - \cos t) + i ( \sin t + \sin t )\right) = \left( 2 e^{it} \right) . $$

Edit(Anton Akhmerov): Yet another statement which could solve the problem is the following: Prove that any $4n\times 4n$ orthogonal matrix $O$ can be brought to a block-diagonal form by a transformation $O\rightarrow S_1 O S_2$ with $S_1$ and $S_2$ symplectic matrices. If this was correct, the rest follows immediately.

  • $\begingroup$ Is your matrix 4n by 4n or actually 2n by 2n? So far the statement looks wrong to me: 1 2 2 3 $\endgroup$ – user11231 Dec 1 '10 at 17:02
  • $\begingroup$ 1) It's 4n by 4n. 2) Your example is not orthogonal, $O^TO \neq 1$. I have checked the statement with random matrices. $\endgroup$ – Anton Akhmerov Dec 1 '10 at 17:06
  • 2
    $\begingroup$ By resorting to a normal form you can reduce to the case of considering $\det(O_{11} - O_{22}) = 0$ only. In the $4 \times 4$-case you get $$\left(\begin{array}{cccc} \cos{\alpha} & -\sin{\alpha} & & \\\ \sin{\alpha} & \cos{\alpha} & & \\\ & & \cos{\beta} & \sin{\beta} \\\ & & \sin{\beta} & -\cos{\beta} \end{array}\right)$$ for which the claim follows from $\sin^{2} + \cos^{2} = 1$. $\endgroup$ – Theo Buehler Dec 1 '10 at 17:36
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    $\begingroup$ The problem is that there is no rotational invariance, so I cannot bring the matrix to a normal form, or at least I don't see how to eliminate offdiagonal blocks. $\endgroup$ – Anton Akhmerov Dec 1 '10 at 17:46
  • $\begingroup$ @Anton: Ooops, sorry, so I need to think about it a little more. $\endgroup$ – Theo Buehler Dec 1 '10 at 17:54

I denote your matrix $\Omega$ by $W$ for the sake of brevity. Note that $W$ is both skew-symmetric and orthogonal, i. e. it satisfies $W=-W^T$ and $W^2=-I$. (And this is all I am going to use about $W$.)

The only thing I am going to use about the matrix $O$ is that $O^TO=I$. The assumption that $O$ is a real matrix will not be needed (it could be from any field of characteristic $\neq 2$).

We are going to use notion of the Pfaffian of a skew-symmetric matrix.

Lemma 1. Let $R$ be a commutative ring with $1$. Let $A\in R^{2n\times 2n}$ be a skew-symmetric matrix, and $B\in R^{2n\times 2n}$ be an arbitrary matrix. Then, the matrix $B^TAB$ is skew-symmetric as well, and $\mathrm{Pf}\left(B^TAB\right)=\det B\cdot\mathrm{Pf} A$.

Proof of Lemma 1. Clearly, the matrix $B^TAB$ is skew-symmetric. It remains to prove that $\mathrm{Pf}\left(B^TAB\right)=\det B\cdot\mathrm{Pf} A$. Let us WLOG assume that $R=\mathbb Q$ (this is WLOG indeed because we are proving a polynomial identity). Since the square of the Pfaffian of a skew-symmetric matrix is the determinant of this matrix, we have

$\left(\mathrm{Pf}\left(B^TAB\right)\right)^2=\det\left(B^TAB\right)=\underbrace{\det B^T}_{=\det B}\cdot\underbrace{\det A}_{=\left(\mathrm{Pf}A\right)^2}\cdot \det B = \left(\det B\cdot\mathrm{Pf} A\right)^2$.`

Thus, FOR EVERY $A$ and $B$, we have either $\mathrm{Pf}\left(B^TAB\right)=\det B\cdot\mathrm{Pf} A$ OR $\mathrm{Pf}\left(B^TAB\right)=-\det B\cdot\mathrm{Pf} A$. By a Zariski-topological argument, we can interchange the words "for every" with the words "or" here, so we obtain: (FOR EVERY $A$ and $B$, we have $\mathrm{Pf}\left(B^TAB\right)=\det B\cdot\mathrm{Pf} A$) OR (FOR EVERY $A$ and $B$, we have $\mathrm{Pf}\left(B^TAB\right)=-\det B\cdot\mathrm{Pf} A$). Since (FOR EVERY $A$ and $B$, we have $\mathrm{Pf}\left(B^TAB\right)=-\det B\cdot\mathrm{Pf} A$) is wrong (take $A=W$ and $B=\mathrm{id}$), we thus must have (FOR EVERY $A$ and $B$, we have $\mathrm{Pf}\left(B^TAB\right)=\det B\cdot\mathrm{Pf} A$), and Lemma 1 is proven.

Lemma 2. Let $R$ be a commutative ring with $1$. Let $B\in R^{2n\times 2n}$ and $C\in R^{2n\times 2n}$ be arbitrary matrices. Then, $\det\left(W-BWC\right)=\det\left(W-CWB\right)$.

Proof of Lemma 2. We have $W-BWC=W\cdot\left(I+WBWC\right)$ (because $-I=W^2$), and thus $\det\left(W-BWC\right)=\det\left(W\cdot\left(I+WBWC\right)\right)=\det W\cdot\det\left(I+WBWC\right)$ and similarly $\det\left(W-CWB\right)=\det W\cdot\det\left(I+WCWB\right)$. Thus, in order to prove Lemma 2, it remains to show that $\det\left(I+WBWC\right)=\det\left(I+WCWB\right)$. This follows from the general fact that if $U$ and $V$ are two square matrices of the same size, then $\det\left(I+UV\right)=\det\left(I+VU\right)$ (for a proof of this fact, apply Corollary 2 in MathLinks post #1491761 to $X=-1$). Thus, Lemma 2 is proven.

Lemma 3. Let $R$ be a commutative ring with $1$. Let $B\in R^{2n\times 2n}$ be an arbitrary matrix. Then, $\mathrm{Pf}\left(W-BWB^T\right)=\mathrm{Pf}\left(W-B^TWB\right)$.

Proof of Lemma 3. Let us WLOG assume that $R=\mathbb Q$ (this is WLOG indeed because we are proving a polynomial identity). Applying Lemma 2 to $C=B^T$, we obtain $\det\left(W-BWB^T\right)=\det\left(W-B^TWB\right)$. Since the determinant of a skew-symmetric matrix is the square of its Pfaffian, this rewrites as $\left(\mathrm{Pf}\left(W-BWB^T\right)\right)^2=\left(\mathrm{Pf}\left(W-B^TWB\right)\right)^2$. Again, a Zariski density argument like in the proof of Lemma 1 shows us that we must have $\mathrm{Pf}\left(W-BWB^T\right)=\mathrm{Pf}\left(W-B^TWB\right)$ (because we cannot have $\mathrm{Pf}\left(W-BWB^T\right)=-\mathrm{Pf}\left(W-B^TWB\right)$ for all $B$, since this would fail for $B=0$). This proves Lemma 3.

Now let us solve our problem:

We have $\mathrm{Pf}U=\mathrm{Pf}\left( -U\right) $ for every skew-symmetric $4n\times 4n$ matrix $U$ (because $\mathrm{Pf}$ is a homogeneous polynomial of degree $2n$). Now,

$\mathrm{Pf}\left( W-OWO^{T}\right) =\mathrm{Pf}\left( W-O^{T}WO\right) $ (after Lemma 3, applied to $B=O$)

$=\mathrm{Pf}\left( -\left( W-O^{T}WO\right) \right) $ (since $\mathrm{Pf}U=\mathrm{Pf}\left( -U\right) $ for every skew-symmetric $4n\times4n$ matrix $U$)

$=\mathrm{Pf}\left( O^{T}WO-W\right) $

$ =\mathrm{Pf}\left( O^{T} WO-O^{T}OWO^{T}O\right) $ (since $O^{T}O=I$ yields $W=O^{T}OWO^{T}O$)

$=\mathrm{Pf}\left( O^{T}\left( W-OWO^{T}\right) O\right) $

$=\underbrace{\det O}_{=-1}\cdot\mathrm{Pf}\left( W-OWO^{T}\right) $ (by Lemma 1, applied to $B=O$ and $A=W-OW^{T}O^{T}$)

$=-\mathrm{Pf}\left( W-OWO^{T}\right) $,

so that $2\mathrm{Pf}\left( W-OWO^{T}\right) =0$ and thus $\mathrm{Pf}\left( W-OWO^{T}\right) = 0$, so that $\det\left( W-OWO^{T}\right) =\left(\mathrm{Pf}\left( W-OWO^{T}\right) \right)^2=0^2=0$, qed.

  • $\begingroup$ Cute ! +1. You could have skipped the proof of Lemma 1, which is classical. But since you need the Zariski density argument, it makes sense to include it. $\endgroup$ – Denis Serre Dec 3 '10 at 10:41
  • $\begingroup$ Thanks! I wanted to include the proof of Lemma 1 in order to refer to it while proving Lemma 3. By the way, do you know a non-Zariski proof of one of these lemmas? $\endgroup$ – darij grinberg Dec 3 '10 at 18:01
  • $\begingroup$ (Of course, this is a very harmless Zariski: All we are actually using is that the polynomials over a field form an integral domain, and that a polynomial over $\mathbb Q$ that is identically zero must be zero as a polynomial. But it's a matter of principle: By forbidding Zariski, we sometimes find novel proofs that show an old result in a new light.) $\endgroup$ – darij grinberg Dec 3 '10 at 18:02

What you are seeing is that the orthogonal matrices of determinant $-1$ swap the two spin representations. The first several parts of this argument will be valid for $(4n+2) \times (4n+2)$ matrices as well, so I'll rename the size of the matrix to $2m$ and specialize to $m$ even when it becomes relevant.

In more detail: Let $S$ be the $2m \times 2m$ change of basis matrix $$\frac{1}{\sqrt{2}} \begin{pmatrix} \mathrm{Id} & i \cdot \mathrm{Id} \\ \mathrm{Id} & -i \cdot \mathrm{Id} \end{pmatrix}$$

Note that $O$ is orthogonal if and only if $SOS^{-1}$ preserves the split bilinear form $x_1 x_{m+1} + x_2 x_{m+2} + \cdots + x_{m} x_{2m}$. Explicitly, $$S O S^{-1} = \frac{1}{2} \begin{pmatrix} O_{11} - i O_{12} + i O_{21} + O_{22} & O_{11} + i O_{12} + i O_{21} - O_{22} \\ O_{11} - i O_{12} - i O_{21} - O_{22} & O_{11} + i O_{12} - i O_{21} + O_{22} \end{pmatrix}$$

We want to show that, for $m$ even: If $U$ preserves the split bilinear form, and has determinant $-1$, then the upper right $m \times m$ block of $U$ is singular.

Consider the action of $U$ on $\bigwedge^{m} \mathbb{C}^{2m}$; this is a big $\binom{2m}{m} \times \binom{2m}{m}$ matrix which we will write $\bigwedge^m U$. The entries of $\bigwedge^m U$ are $m \times m$ minors of $U$. In particular, the entry we care about is the coefficient with which $e_1 \wedge e_2 \wedge \cdots \wedge e_m$ is taken to $e_{m+1} \wedge e_{m+2} \wedge \cdots e_{2m}$.

Let $\alpha_{+}$ and $\alpha_{-}$ be the highest weights of the spin reps of $\mathrm{Spin}(2m)$. Let $V_{\pm}$ be the irrep with highest weight $2 \alpha_{\pm}$. Then $V_{+}$ and $V_{-}$ both inject into $\bigwedge^{m} \mathbb{C}^{2m}$. The vector $e_1 \wedge \cdots \wedge e_m$ is a high weight vector for $V_{+}$. The vector $e_{m+1} \wedge \cdots e_{2m}$ is a low weight vector for $V_{(-1)^m}$. In particular, when $m$ is even, Both these wedge products lie in $V_{+}$. An orthogonal matrix with determinant $-1$ switches the spin representations. So this entry of $\bigwedge^m U$ is $0$.

If you'd like a "geometric" presentation of the same argument, note that $$O_{11} + i O_{12} + i O_{21} - O_{22} = \begin{pmatrix} \mathrm{Id} & i \cdot \mathrm{Id} \end{pmatrix} O \begin{pmatrix} \mathrm{Id} \\ i \cdot \mathrm{Id} \end{pmatrix}.$$

The kernel of the left hand matrix is an isotropic $m$-plane; call it $L$. The image of the right hand matrix is also $L$. So we want to show that, if $O$ has determinant $-1$, then $L \cap O L$ is nonzero.

Let $Q$ be the quadric hypersurface $\sum x_i^2=0$ in $\mathbb{CP}^{2m-1}$. This is a complex variety of (complex) dimension $2m-2$. The projectivization $\mathbb{P} L$ is a subvariety of dimension $m-1$. It is well known that $H^{2m-2}(Q) \cong \mathbb{Z}^2$, with orthogonal matrices of determinant $1$ switching the two generators. It is less well known that the intersection form on $H^{2m-2}(Q)$ has matrix $\left( \begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix} \right)$ when $m$ is even and ` $\left( \begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix} \right)$ when $m$ is odd.

We are in the former case. So the cup product $[\mathbb{P}L] \cup [O \cdot \mathbb{P}L]$ is $1$ and we deduce that $L$ and $OL$ have nonzero intersection, as desired.

  • $\begingroup$ how is this not the accepted answer? $\endgroup$ – Vivek Shende Dec 2 '10 at 10:57
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    $\begingroup$ This answer is great, and I'm very thankful to David but unfortunately my math education is insufficient to understand it fully. So I accepted the other answer because I can understand and reproduce it myself. I'd accept two answers, was it possible. $\endgroup$ – Anton Akhmerov Dec 2 '10 at 15:04
  • $\begingroup$ I fixed some over simplifications in the representation theory. $\endgroup$ – David E Speyer Dec 2 '10 at 18:22

Might be getting there. So far, I have

$$ \begin{array}{ll} & \left(O_{11} - O_{22} + i O_{12} + i O_{21} \right) \cdot \left( O_{11}^t + O_{22}^t - i O_{12}^t + i O_{21}^t \right) = \\\ & (O_{11} O_{22}^t - O_{22} O_{11}^t ) + (O_{21} O_{12}^t - O_{12} O_{21}^t ) + \\\ & i (O_{12} O_{11}^t - O_{11} O_{12}^t ) + i (O_{21} O_{22}^t - O_{22} O_{21}^t ) \end{array}$$ so that the resulting real and imaginary parts are skew symmetric. However, as they are $ 2n \times 2n$ this does not imply singular for either part.

Note: there are some generally unknown but useful properties of orthogonal matrices, for example, in this case: $$ \det O_{11} = - \det O_{22}, $$ see

What's your favorite equation, formula, identity or inequality?

  • $\begingroup$ Where did the term $O_{11} O_{11}^T$ and the similar ones go? $\endgroup$ – Anton Akhmerov Dec 1 '10 at 18:58
  • $\begingroup$ Cancellation, identity matrix minus another identity matrix. $\endgroup$ – Will Jagy Dec 1 '10 at 18:59

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