Why can't the $\mathbb{Z}_2$ invariant be defined with a determinant instead of a pfaffian?

Question

I'm reading through Topological Insulators and Topological Superconductors by Bernevig and Hughes. I'm on Chapter 10, where he describes the original formulation of the $\mathbb{Z}_2$ invariant given in the original Kane-Mele paper. Both expositions are basically identical.

They consider an $N$-band model, and they denote the $N$-component wavefunction in $k$-space as $|u_k^n\rangle$, $n=1,..., N$. They define the quantity $P(k)=\text{Pf}(\langle u_k^n|\hat T|u_k^m\rangle)$, where $\hat{T}$ is the time-reversal operator. Then they argue the following things are true of $P(k)$:

• $|P(k)|=1$ if $k$ is a time-reversal-invariant momentum. This is because the matrix $\langle u_k^n|\hat T|u_k^m\rangle$ is unitary at such a point.
• If the phase of $P(k)$ winds around a loop, there is some $k_0$ inside that loop with $P(k_0)=0$. This is because $P(k)$ is a continuous function of $k$.
• If the phase of $P(k)$ winds $z$ times around $k_0$, then it winds $-z$ times around $-k_0$. This is because of time-reversal stuff.

The combination of those three facts makes them say that the total winding around the zeros of the $k_x>0$ half of the Brilloiun zone (mod 2) is a topological invariant. They argue that if we have a single vortex in the right half of the Brilloiun zone, the only way it can disappear is by combining with it's time reversed partner vortex in the left half of the Brilloiun zone. But because of point (3) above, the only place these vortexes can meet is at a time-reversal-invariant point, which is impossible because of (1).

I can accept all of that. But I don't understand: Why introduce the Pfaffian at all? The normal determinant should have all of those properties too! Since the determinant is just the pfaffian squared, it should be zero at the same time, it should have absolute value 1 at the same time, and if one winds the other should wind (twice as much, but still)! Why not just define $D(k)=\text{det}(\langle u_k^n|\hat T|u_k^m\rangle)$, and define your topological invariant to be the number of times $D(k)$ winds over half the Brillouin zone? You can just divide by two at the end and you get back the same answer, no? And who doesn't prefer determinants to pfaffians?

What am I missing here?

Answer 1

This is not my specialty, but the Bernevig and Hughes text looked interesting, so I spent some time with it.

I might be missing something, but as far as I can tell from the pieces I've digested, you're right. For the vorticity-based construction of the $Z_2$ invariant, the Pfaffian is being used only in what is presumably a connected region where it is continuous and nonzero (namely the boundary of the half-BZ); and in any such region, the Pfaffian can be reconstructed from its square except for an overall sign that has no bearing on the vorticity.

However, section 10.6 in the Bernevig and Hughes text highlights something that might help explain the general preference for using Pfaffians in this context, in addition to the reasons that were already mentioned in other comments/answers. Section 10.6 reviews two different ways of constructing the same $Z_2$ invariant, using the Pfaffians of two different quantities at different sets of points. Specifically, they define one matrix $m(\mathbf{k})$ that is antisymmetric for all $\mathbf{k}$ in the Brillouin zone (BZ), and they define another matrix $B(\mathbf{k})$ that is antisymmetric only at time-reversal invariant points. According to section 10.6, the same $Z_2$ invariant may be expressed in either of two ways:

• The modulo-$2$ vorticity of the Pfaffian of $m$ in half the BZ. This seems to be the construction used in the original Kane-Mele paper.

• The product of the signs of the Pfaffian of $B$ at those points where $B$ is antisymmetric, where the sign is defined by comparing it to one consistent branch of the square-root of the determinant. In this construction, knowing only the determinant of $B$ at those same points is clearly not sufficient.

These two constructions of the $Z_2$ invariant are also reviewed in "Notes on topological insulators," https://arxiv.org/abs/1501.02874. This paper and section 10.4 in the Bernevig and Hughes text both indicate that the second construction, the product-of-signs version, can also be applied in higher dimensions (not just 2d).

So instead of this:

And who doesn't prefer determinants to Pfaffians?

maybe Kane and Mele were thinking this:

And who doesn't prefer Pfaffians to determinants?

...because sometimes it does matter, and when it does, the Pfaffian is more informative.

Answer 2

FWIW, I agree with you.

The index is

$$\frac{1}{2\pi i} \int_C d \vec k \cdot \nabla \log(P(k) + i \epsilon),$$

where $P(k)$ is the Pfaffian. The determinant is $D(k) = P(k)^2$ so we have

$$\log(D(k) + i2\epsilon) = \log((P(k) + i\epsilon)^2) = 2\log(P(k)+i\epsilon),$$

so the index is also

$$\frac{1}{4\pi i} \int_C d \vec k \cdot \nabla \log(D(k) + i \epsilon).$$

I think Kane and Mele might also agree with you. From their paper:

I wanted to point out that the fermionic ground state here is a Slater determinant, which is the point of the formula in the snippet for $P(\alpha)^2$. The determinant shows up because if you exchange indices of a matrix, ie. multiplying by a transposition matrix, the determinant picks up a minus sign. Thus, determinants are natural things for expressing fermionic ground states.

On the other hand, Pfaffians show up when computing (many-body) partition functions of real fermions. If we were computing some twisted partition function of this model, then we really would be concerned with its sign, since that would be precisely this $\mathbb{Z}_2$ invariant. It's an interesting general feature that winding invariants of single-particle ground states are captured by the many-body partition function. This is how the Pfaffian shows up in this classic (where you really need the Pfaffian and not the determinant), and I assume Kane and Mele had this on their mind when they wrote about the Pfaffian, see also Chiral Anomaly's answer. That square-root in the snippet looks reaaally spurious to me.

Answer 3

Pfaffians are not weird objects but very natural for fermionic systems. They follow very similar rules as determinants.

You are missing the topological information in the sign of the Pfaffian. Taking a square root changes the topology of the manifold on which the function is single valued - just like the complex square root is properly defined on a topologically nontrivial Riemannian surface and not just on the plane of complex numbers.

Because of periodicity, winding over a Brillouin zone is well-defined but winding over half a Brillouin zone is path-dependent (without additional arguments).

Answer 4

TL;DR:

1. Firstly, the main point is that one would always want to define & use the most fundamental/elementary objects in a theory, in OP's case, the Pfaffian (F), even if it doesn't matter in specific calculations, such as, e.g., the index $$\begin{align}\frac{1}{2\pi i}&\oint_{\frac{1}{2}{\rm BZ}} \! \mathrm{d}{\rm Ln}\{{\rm Pf}(A(k))+i0^+\} \cr~=~& \frac{1}{4\pi i}\oint_{\frac{1}{2}{\rm BZ}} \! \mathrm{d}{\rm Ln}\{{\rm Det}(A(k))+i0^+\},\end{align} \tag{A}$$ which OP is asking about, cf. eq. (5) in Kane & Mele, and Ryan Thorngren's answer.

   Why potentially lose information by working with composite objects? This is also the main point of Chiral Anomaly's answer.

2. Secondly in this answer, we want to stress the transformation properties (I) of the determinant (G), which somehow highlights its composite nature (H), in contrast to a fundamental/elementary determinant (C), cf. eqs. below.

After this preample, let's provide details.

1. On one hand, let $V$ be an $n$-dimensional $\mathbb{C}$-vector space with a basis $(e_j)_{j=1, \ldots, n}$. Let $V^{\ast}$ be the dual vector space with the dual basis $(e^{\ast k})_{k=1, \ldots, n}$. Let $$T~=~e_j~ T^j{}_k\otimes e^{\ast k}~ \in~V\otimes V^{\ast}~\cong~{\cal L}(V;V)\tag{B}$$ be a linear map from $V$ to $V$. Then the notion of determinant $${\rm Det} (T)~:=~{\rm Det} (T^j{}_k)\tag{C}$$ is invariant under a change of basis $$ e^{\prime}_k~=~e_j M^j{}_k, \qquad T^{\prime i}{}_{\ell}~=~M^i{}_j T^j{}_k(M^{-1})^k{}_{\ell}. \tag{D}$$

2. On the other hand, consider a 2-form $$A~=~\frac{1}{2}e_j~ A^{jk}\wedge e_k~ \in~V\wedge V\tag{E}$$ with an antisymmetric matrix $A^{jk}$. The Pfaffian can e.g. be defined via a Grassmann-odd Gaussian integral representation as$^1$ $$ {\rm Pf}(A)~:=~\int \!d\theta_n \ldots d\theta_1~ e^{\frac{1}{2}\theta_j A^{jk}\theta_k}.\tag{F}$$ Similarly it is not hard to show that the Grassmann-odd Gaussian integral representation $${\rm Det}(A)~:=~\int \!d\theta_1 ~d\widetilde{\theta}_1 \ldots d\theta_n ~d\widetilde{\theta}_n~ e^{\widetilde{\theta}_j A^{jk}\theta_k} \tag{G}$$ reproduces the standard definition of a determinant, and that $$ {\rm Det}(A)~=~{\rm Pf}(A)^2,\tag{H} $$ cf. e.g. my Math.SE answer here.

In contrast to the determinant (C), the Pfaffian (F) and the determinant (G) are not invariant, but transform as $$\begin{align}{\rm Pf}(A^{\prime})~=~&{\rm Det}(M){\rm Pf}(A),\cr {\rm Det}(A^{\prime})~=~&{\rm Det}(M)^2{\rm Det}(A),\end{align}\tag{I}$$ under a change of basis $$ e^{\prime}_k~=~e_j M^j{}_k, \qquad A^{\prime i\ell}~=~M^i{}_j A^{jk}(M^t)_k{}^{\ell}, \tag{J}$$ cf. e.g. this Math.SE post.

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$^1$ We use the sign convention that Berezin integration $$\int d\theta_i~\equiv~\frac{\partial}{\partial \theta_i}\tag{K} $$ is the same as differentiation wrt. $\theta_i$ acting from left. See e.g. this Phys.SE post.