Lecture 2: Spectral partitioning and near-optimal foams

In the last lecture, we reduced the problem of cheating in \mathcal G_m^{\otimes k} (the k-times repeated m-cycle game) to finding a small set of edges \mathcal E in (\mathbb Z_m^k)_\infty whose removal eliminates all topologically non-trivial cycles.  Such a set \mathcal E is called a spine. To get some intuition about how many edges such a spine should contain, let’s instead look at a continuous variant of the problem.

Spines, Foams, and Isoperimetry

Consider again the k-dimensional torus \mathcal T^k = \mathbb R^k/\mathbb Z^k, which one can think of as \lbrack 0,1)^k with opposite sides identified.  Say that a nice set (e.g. a compact, C^\infty surface) \mathcal E \subseteq \mathcal T^k is a spine if it intersects every non-contractible loop in \mathcal T^k.  This is the continuous analog of a spine in (\mathbb Z_m^k)_\infty.  We will try to find such a spine \mathcal E with surface area, i.e. \mathrm{Vol}_{k-1}(\mathcal E), as small as possible.

Let’s consider some easy bounds.  First, it is clear that the set

\displaystyle \mathcal E = \left\{(x_1, \ldots, x_k) \in [0,1)^k : \exists i \in \{1,2,\ldots,k\}, x_i = 0\right\}

is a spine with \mathrm{Vol}_{k-1}(\mathcal E) = k.  (A moment’s thought shows that this is “equivalent” to the provers playing independent games in each coordinate of \mathcal G_m^{\otimes k}.)

To get a good lower bound, it helps to relate spines to foams which tile \mathbb R^k according to \mathbb Z^k, as follows.  Take two potential spines.

To determine which curve is actually a spine, we can repeatedly tile them side-by-side.

The first tiling contains the blue bi-infinite curve, which obviously gives a non-trivial cycle in \mathcal T^k, while the second yields a tiling of \mathbb R^k by bodies of volume 1.  It is easy to deduce the following claim.

Claim: A surface \mathcal E \subseteq \mathcal T^k is a spine if and only if it induces a tiling of the plane by bodies of volume 1 which is invariant under shifts by \mathbb Z^k.

By the isoperimetric inequality in \mathbb R^k, this immediately yields the bound

\displaystyle \mathrm{Vol}_{k-1}(\mathcal E) \geq \mathrm{Vol}_{k-1}(S^{k-1}) \approx \sqrt{k},

where S^{k-1} is the unit (k-1)-dimensional sphere.

So the the surface area of an optimal spine lies somewhere between k and \sqrt{k}.  On the one hand, cubes tile very nicely but have large surface area.  On the other hand, we have sphere-like objects which have small surface area, but don’t seem (at least intuitively) to tile very well at all.  As first evidence that this isn’t quite right, note that it is known how to cover \mathbb R^k by disjoint bodies of volume at most 1 so that the surface area/volume ratio grows like \sqrt{k}.  See Lemma 3.16 in this paper, which is based on Chekuri, et. al.  It’s just that these covers are not invariant under \mathbb Z^k shifts.

Before we reveal the answer, let’s see what consequences the corresponding discrete bounds would have for parallel repetition of the m-cycle game.  If the “cube bound” were tight, we would have \mathsf{val}(\mathcal G_m^{\otimes k}) \approx 1 - \frac{k}{m}, which doesn’t rule out a strong parallel repetition theorem (\alpha^*=1 in the previous lecture).  If the “sphere bound” were tight, we would have \mathsf{val}(\mathcal G_m^{\otimes k}) \approx 1 - \frac{\sqrt{k}}{m}, which shows that \alpha^* \geq 2.   In the latter case, the approach to proving equivalence of the UGC and MAX-CUT conjectures doesn’t even get off the ground.

As the astute reader might have guessed, recently Ran Raz proved that \mathsf{val}(\mathcal G_m^{\otimes k}) \geq 1 - C\frac{\sqrt{k}}{m} for some constant C > 0, showing that a strong parallel repetition theorem—even for unique games—is impossible.   Subsequently, Kindler, O’Donnell, Rao, and Wigderson showed that there exists a spine \mathcal E \subseteq \mathcal T^k with \mathrm{Vol}_{k-1}(\mathcal E) \approx \sqrt{k}.  While it is not difficult to show that the continuous result implies Raz’s discrete result, we will take a direct approach found recently by Alon and Klartag.

Spectral partitioning, Cheeger’s inequality, and Dirichlet boundary conditions

First, we will show that it suffices to find a subset W \subseteq \mathbb Z_m^k so that the induced graph on W contains no non-trivial cycles, and E(W, \overline{W}) is small, where we use E(S, T) to denote the set of edges from S to T in (\mathbb Z_m^k)_\infty.  A variant of this reduction appears in KORW.  For a subset W \subseteq Z_m^k, let h(W) = \frac{|E(W,\overline{W})|}{|W|}.

Random Partitioning Lemma: For any subset W \subseteq Z_m^k which contains no non-trivial cycles, there exists a spine \mathcal E \subseteq (\mathbb Z_m^k)_\infty with |\mathcal E| \leq h(W) m^k.


Let x_1, x_2, \ldots \in \mathbb Z_m^k be i.i.d. uniformly random vectors, and put W_i = x_i + W = \{ x_i + w : w \in W \}, with addition done over \mathbb Z_m^k.  Clearly (with probability 1), we have \mathbb Z_m^k = \bigcup_{i=1}^t W_i for some finite time t.  Let us define C_i = W_i \setminus \bigcup_{j < i} W_j, so that \mathbb Z_m^k = \bigcup_{i=1}^t C_i is a partition into disjoint sets.  Since each W_i is isomorphic to W, and C_i \subseteq W_i, no set C_i contains a non-trivial cycle.  So if we define \mathcal E = \bigcup_{i \neq j} E(C_i, C_j), then we certainly get a spine.

To calculate \mathbb E\,|\mathcal E|, we will use a “charging” argument common to many random clustering analyses.  If u \in C_i, then we put \mathcal C(u) = E(u, \cup_{j > i} C_j).  Clearly we can write \mathcal E = \bigcup_{u \in \mathbb Z_m^k} \mathcal C(u), so it suffices to estimate \mathbb E|\mathcal C(u)| for a fixed u.

To this end, note that after conditioning on u \in C_i, u is a uniformly random vertex of W_i, and thus

\displaystyle \mathbb E|\mathcal C(u)| \leq \mathbb E_{C_i}\mathbb E\left[|E(u, \overline{W_i})|\,:\, u \in C_i\right] = \mathbb E_{C_i} \frac{|E(W_i, \overline{W_i})|}{|W_i|} = h(W).

By linearity of expectation, it follows that \mathbb E|\mathcal E| \leq h(W) m^k.

So our task is reduced to finding a subset W with h(W) small and which contains no non-trivial cycles.  A natural method for finding such a set W would be to prove a bound on the smallest non-zero eigenvalue of the Laplacian on (\mathbb Z_m^k)_\infty, and then use Cheeger’s inequality to conclude the existence of a good cut.  But here we have to handle the special condition that W should contain no non-trivial cycles.

There is a very natural way to ensure this:  If we impose that our eigenfunction vanishes on the boundary of \{0,1,\ldots,m-1\}^k (of course any other fundamental domain would do), then the standard “sweep” algorithm which chooses a level set of the eigenfunction will always find a set W that doesn’t wrap around (and therefore doesn’t contain any non-trivial cycles).  The orange lines are examples of possible level sets.

Dirichlet boundary conditions for the Laplacian.

Fix a graph G=(V,E) with V = \{1,2,\ldots,n\}.  We define its combinatorial Laplacian as the matrix \Delta = D - A, where D is the diagonal degree matrix with D_{ii} = \mathrm{deg}(i), and A is the adjacency matrix of G.  Given a subset S \subseteq V, we can consider the first Dirichlet eigenvalue with boundary conditions on S:

\displaystyle \lambda_1^S = \min_{\substack{0 \neq v \in \mathbb R^n \\ v_i = 0, i \in S}} \frac{\langle v, \Delta v\rangle}{\langle v, v \rangle}.

By standard variational principles, there always exists a non-negative vector v \in \mathbb R^n with v_i = 0 for i \in S and (\Delta v)_i = \lambda_1^S v_i for i \in V \setminus S.  If we take \|v\|=1, then this is the lowest-energy norm-1 function on G, subject to the boundary conditions.

Next, we need a version of Cheeger’s inequality for \lambda_1^S.  As usual, we won’t actually need an eigenvector—any vector that has small Rayleigh quotient will do.

Discrete Cheeger inequality: If v_i = 0 for all i \in S, then

\displaystyle \frac{\langle v, \Delta v\rangle}{\langle v,v \rangle} \geq \frac{1}{2d_{\max}} \min_{W \subseteq V \setminus S} h(W)^2

where d_{\max} is the maximum degree in G.

For the proof, I’ll refer to Theorem 4 in Alon-Klartag.  Note that the proof for the Dirichlet version is even simpler than for the “standard” (Neumann) eigenvalues—it’s just a straightforward combination of Cauchy-Schwarz and the coarea formula.

Choosing a good eigenvector.

So to finish cheating at the odd-cycle game, we need only produce a low-energy vector on (\mathbb Z_m^k)_\infty.  We are aided greatly by the fact that this graph is a product of m-cycles, and thus we will only need to know about eigenvectors on the m-cycle (\mathbb Z_m)_\infty.

Let A be the adjacency matrix of the m-cycle, and let A' be the matrix obtained from it by replacing the last row and column by the zero vector (thus A' is the adjacency matrix of the (m-1)-path, with an isolated vertex).  Now, the adjacency matrix of (\mathbb Z_m^k)_\infty is simply (I+A)^{\otimes k} - I^{\otimes k}, where (\cdot)^{\otimes k} denotes the k-fold tensor product of a matrix with itself.  Note that if v \in \mathbb R^m satisfies v_m = 0, then

\displaystyle \langle v^{\otimes k}, [(I+A)^{\otimes k} - I^{\otimes k}] v^{\otimes k} \rangle = \langle v^{\otimes k}, [(I+A')^{\otimes k} - I^{\otimes k}] v^{\otimes k} \rangle\quad (1)

It’s easy to see that the vector v \in \mathbb R^m with v_i = \sin(\pi\frac{i}{m}) is an eigenvector of A' with eigenvalue \lambda = 2 \cos(\pi/m) and v_m=0. Therefore v^{\otimes k} is an eigenvector of (I+A')^{\otimes k} - I^{\otimes k} with eigenvalue (1+\lambda)^k-1.  Using (1), this implies that

\displaystyle \langle v^{\otimes k}, [(I+A)^{\otimes k} - I^{\otimes k}] v^{\otimes k} \rangle = [(1+\lambda)^k-1] \|v^{\otimes k}\|^2

Finally, using the fact that the Laplacian of (\mathbb Z_m^k)_\infty is \Delta = (3^k-1) I^{\otimes k} - [(I+A)^{\otimes k}-I^{\otimes k}], we see that

\displaystyle \frac{\langle v^{\otimes k}, \Delta v^{\otimes k}\rangle}{\langle v, v \rangle} = 3^k - (1+\lambda)^k \approx 3^k \frac{k}{m^2},

for k \ll m^2.

Applying the Discrete Cheeger Inequality above, with S = \{ x \in \mathbb Z_m^k : \exists i\,x_i \equiv 0\,\mathrm{mod}\ m \} implies that there exists a W \subseteq \mathbb Z_m^k \setminus S, with

\displaystyle h(W) \lesssim 3^k \frac{\sqrt{k}}{m}.

Since W \cap S = \emptyset, W contains no non-trivial cycles.  Applying the Random Partitioning Lemma, and noting that the number of edges in \mathbb Z_m^k is \frac12 (3^k-1) m^k, we conclude that there exists a spine which cuts at most a

\displaystyle \delta(k,m) \lesssim \frac{\sqrt{k}}{m}

fraction of edges.

As discussed above, this finishes the proof that \alpha = 2 is the best possible exponent in the parallel repetition theorem (even for unique games), and thus this kind of gap amplification fails to mount an attack on the Unique Games Conjecture.  In the next lecture, we’ll see a more benevolent use of eigenvectors.

5 thoughts on “Lecture 2: Spectral partitioning and near-optimal foams

  1. Just a minor remark about the statement: “If the “sphere bound” were tight, […] the approach to proving equivalence of the UGC and MAX-CUT conjectures doesn’t even get off the ground.”

    I guess, this is not quite accurate. The point is that the odd cycle is not a hard instance of MAX-CUT. So, Raz + KORW don’t rule out the approach you sketched in your first lecture. It could have still been the case that strong parallel repetition were true on a class of interesting (i.e. hard) instances of MAX-CUT. But you may have addressed this issue in your third lecture :-)

  2. Hi Moritz,
    The approach of Lecture 1 (that of Feige-Kindler-O’Donnell) was to prove a general parallel repetition theorem for unique games with exponent < 2, and use this to show that the MAX-CUT conjecture implies the UGC. Certainly this is ruled out by Raz’s example. As you say, it could still be the case that “hard instances” of MAX-CUT perform better under repetition. This will be ruled out in Lecture 3, via your SDP approach, but Lecture 3 is a bit delayed because I want to rework the presentation.

  3. hello,

    I know this is not the right place but I don’t know where to put my question
    I heard a story about a student who slept in a lecture and the he woke up to find that the prof. wrote 2 math problems on the board and he thought that these two questions are homework so he tried to solve them and he solved only one, then he gave his answer to the prof. who said that this wasn’t a homework but it was a problem that scientest had NEVER proved it .. !!!

    but the student solved it! because he tried his best (not expecting that this problem can’t be solved !)

    is this a TRUE story ??

    sorry for bothering you prof. .. but I don’t know where to ask as I mentioned above

    Abdullah Abukar

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