Tag Archives: topology

Calabi-Yau Manifolds and Moduli Stabilization

For better or worse, string theory dominates modern research in theoretical physics. Naively, you might expect a theory consisting of tiny strings to be pretty simple. But the subject has grown into a vast and exciting playground for new ideas.

String theory is popular not just because it might unify physics or quantize gravity. In fact, many unexpected offshoots have proved more successful than the original idea! From particle physics to superconductors, string theory is having a surprising indirect impact. It’s certainly useful, even if it doesn’t prove to be the ultimate description of reality.

But what of the original plan – to describe nature using strings? A key sticking point is the existence of extra dimensions. String theory needs 6 of these to work consistently. Another problem is supersymmetry. 10 dimensional string theory must have lots of this to work correctly. But 4 dimensional physics only has a little bit of supersymmetry at most!

It turns out that these problems can be solved in one step. By coiling up the extra dimensions into a Calabi-Yau manifold we can make the extra dimensions effectively invisible, while reducing the supersymmetry we end up with in our 4D world. So what is this Calabi-Yau manifold, I hear you ask!

Well, Calabi-Yau is just a technical term for the shape of the compact extra dimensions. Different shapes break different amounts of symmetries, leaving us with different theories. Calabi-Yau’s are just symmetrical enough to break the right amount of supersymmetry, giving us a sensible theory in the end!

Technically Calabi-Yau manifolds must have a metric which is Kaehler and Ricci flat. These properties provide the correct information about the shape of the curled up dimensions. So we must look for 6-real dimensional manifolds with these properties.

Generically, you don’t have to put a notion of distance on a space. When I go for a walk, I don’t always carry around a yardstick so I can measure how far I’ve gone! You can have a perfectly good manifold without giving it a metric, but you get extra information once you have defined what distance means.

As it happens, finding a metric which is Calabi-Yau is quite difficult. But due to the genius of Shing-Tung Yau, we know that you don’t need to do this! There’s an equivalent definition of a Calabi-Yau manifold which doesn’t depend on metrics at all. All you need to know is the topological information about the manifold – roughly speaking, how “holey” it is.

If you know something about differential geometry, this kind of equivalence might sound familiar. Yau’s theorem relating geometry and topology is like a (much) more complicated version of the classic Gauss-Bonnet theorem!

It’s a darn sight easier to discover Calabi-Yau’s when you know it’s only the topological data that matters. At first people thought there might only be a few, but now we know there’s a huge number of potential candidates! The problem then becomes choosing one which produces the physics of our universe.

While people have made progress on this, the going is tough. One reason is that nobody knows the metric on a compact Calabi-Yau. This isn’t so important for string calculations, but it makes a big difference when you need to consider branes. So people have come up with various workarounds, which give promising physical results. One such success story is provided by my colleague Zac Kenton, who recently wrote a paper on brane inflation with his PhD supervisor, Steve Thomas.

There’s one final complication that I should mention. If string theory is to be a fundamental theory, then the Calabi-Yau shape should be dynamic. More specifically it will squeeze and stretch over time, unless there’s some mechanism to keep it stable. From the perspective of the 4 large dimensions, this freedom is seen as free scalar fields. These so-called “moduli” fields are bad, because we don’t observe anything like them in nature!

To solve this problem, we must find a way of constraining the fluctuations of the Calabi-Yau. Put another way we have to stabilize the moduli fields, by giving them potential terms, so that their fluctuations are small and essentially negligible at low energies. Hence this is known as the problem of moduli stabilization.

One popular way to solve the conundrum is to turn on some supergravity fields at high energy. These so-called fluxes generate potential terms for the moduli, solving the stabilization problem. Initially this idea was unpopular because of a famous no-go theorem by Witten. But since the advent of the D-brane revolution, the concept is back in vogue!

So there you have it – a 5 minute snapshot of “real” string theory. Now it’s time to get back to my calculations, where string theory is more the background Muse, and certainly not the main protagonist!

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Algebra, Geometry and Topology: An Excellent Cocktail

Yes and I’ll have another one of those please waiter. One shot Geometry, topped up with Algebra and then a squeeze of Topology. Shaken, not stirred.

Okay, I admit that was both clichéd and contrived. But nonetheless it does accurately sum up the content of this post. We’ll shortly see that studying affine varieties on their own is like having a straight shot of gin – a little unpleasant, somewhat wasteful, and not an experience you’d be keen to repeat.

Part of the problem is the large number of affine varieties out there! We took a look at some last time, but it’s useful to have just a couple more examples. An affine plane curve is the zero set of any polynomial in \mathbb{A}^2. These crop up all the time in maths and there’s a lot of them. Go onto Wolfram Alpha and type plot f(x,y) = 0 replacing the term f(x,y) with any polynomial you wish. Here are a few that look nice

f(x,y) = y^2 – x^2 – x^3
f(x,y) = y^3 – x – x^2y
f(x,y) = x^2*y + x*y^2-x^4-y^4

There’s a more general notion than an affine plane curve that works in \mathbb{A}^n. We say a hypersurface is the zero of a single polynomial in \mathbb{A}^n. The cone in \mathbb{R}^3 that we say last time is a good example of a hypersurface. Finally we say a hyperplane is the zero of a single polynomial of degree 1 in \mathbb{A}^n.

Hopefully all that blathering has convinced you that there really are a lot of varieties, and so it looks like it’s going to be hard to say anything general about them. Indeed we could look at each one individually, study it hard and come up with some conclusions. But to do this for every single variety would be madness!

We could also try to group them into different types, then analyse them geometrically. This way is a bit more efficient, and indeed was the method of the Ancients when they learnt about conic sections. But it is predictably difficult to generalise this to higher dimensions. Moreover, most geometrical groupings are just the tip of the iceberg!

What with all this negativity, I imagine that a shot of gin sounds quite appealing now. But bear with me one second, and I’ll turn it into a Long Island Iced Tea! By broadening our horizons a bit with algebraic and topological ideas, we’ll see that all is not lost. In fact there are deep connections that make our (mathematical) life much easier and richer, thank goodness.

First though, I must come good on my promise to tell you about some subset’s of \mathbb{C}^n that aren’t algebraic varieties. A simple observation allows us to come up with a huge class of such subsets. Recall that polynomials are continuous functions from \mathbb{C}^n to \mathbb{C}, and therefore their zero sets must be closed in the Euclidean topology. Hence in particular, no open ball in \mathbb{C}^n can be thought of as a variety. (If you didn’t understand this, it’s probably time to brush up on your topology).

There are two further ‘obvious’ classes. Firstly graphs of transcendental functions are not algebraic varieties. For example the zero set of the function f(x,y) = e^{xy}-x^2 is not an affine variety. Secondly the closed square \{(x,y)\in \mathbb{C}^2:|x|,|y|\leq 1\} is an example of a closed set which is not an affine variety. This is because it clearly contains interior points, while no affine variety in \mathbb{C}^2 can contain such points. I’m not entirely sure at present why this is, so I’ve asked on math.stackexchange for a clarification!

How does algebra come into the mix then? To see that, we’ll need to recall a definition about a particular type of ring.

Definition 2.1 A Noetherian ring is a ring which satisfies the ascending chain condition on ideals. In other words given any chain I_1 \subseteq I_2 \subseteq \dots \ \exists n s.t. I_{n+k}=I_n for all k\in\mathbb{N}.

It’s easy to see that all fields are trivially Noetherian, for the only ideals in k are $latex 0$ and k itself. Moreover we have the following theorem due to Hilbert, which I won’t prove. You can find the (quite nifty) proof here.

Theorem 2.2 (Hilbert Basis) Let N be Noetherian. Then N[x_1] is Noetherian also, and by induction so is N[x_1,\dots,x_n] for any positive integer n.

This means that our polynomial rings k[\mathbb{A}^n] will always be Noetherian. In particular, we can write any ideal I\subset k[\mathbb{A}^n] as I=(f_1, \dots, f_r) for some finite r, using the ascending chain condition. Why is this useful? For that we’ll need a lemma.

Lemma 2.3 Let Y be an affine variety, so Y=V(T) some T\subset K[\mathbb{A}^n]. Let J=(T), the ideal generated by T. Then Y=V(J).

Proof By definition T\subset J so V(J)\subset V(T). We now need to show the reverse inclusion. For any g\in J there exist polynomials t_1,\dots, t_n in T and q_1,\dots,q_n in K[\mathbb{A}^n] s.t. g=\sum q_i t_i. Hence if p\in V(T) then t_i(p)=0 \ \forall i so p\in V(J). \blacksquare

Let’s put all these ideas together. After a bit of thought, we see that every affine variety Y can be written as the zero set of a finite number of polynomials t_1, \dots,t_n. If you don’t get this straight away look back carefully at the theorem and the lemma. Can you see how to marry their conclusions to get this fact?

This is an important and already somewhat surprising result. If you give me any subset of \mathbb{A}^n obtained from the solutions (possibly infinite) number of polynomial equations, I can always find a finite number of equations whose solutions give your geometrical shape! (At least in theory I can – doing so in practice is not always easy).

You can already see that a tiny bit of algebra has sweetened the cocktail! We’ve been able to deduce a fact about every affine variety with relative ease. Let’s pursue this link with algebra and see where it takes us.

Definition 2.4 For any subset X \subset \mathbb{A}^n we say the ideal of X is the set I(X):=\{f \in k[\mathbb{A}^n] : f(x)=0\forall x\in X\}.

In other words the ideal of X is all the polynomials which vanish on the set X. A trivial example is of course I(\mathbb{A}^n)=(0). Try to think of some other obvious examples before we move on.

Let’s recap. We’ve now defined two maps V: \{\textrm{ideals in }k[\mathbb{A}^n]\}\rightarrow \{\textrm{affine varieties in }\mathbb{A}^n\} and I:\{\textrm{subsets of }\mathbb{A}^n\}\rightarrow \{\textrm{ideals in }k[\mathbb{A}^n]\}. Intuitively these maps are somehow ‘opposite’ to each other. We’d like to be able to formalise that mathematically. More specifically we want to find certain classes of affine varieties and ideals where V and I are mutually inverse bijections.

Why did I say certain classes? Well, clearly it’s not the case that V and I are bijections on their domains of definition. Indeed V(x^n)=V(x), but (x)=\neq(x^n) so V isn’t injective. Furthermore working in \mathbb{A}^1_{\mathbb{C}} we see that I(\mathbb{Z})=(0)=I(\mathbb{A}^1) so I is not injective. Finally for n\geq 2 \ (x^n)\notin \textrm{Im}(I) so I is not surjective.

It’ll turn out in the next post that a special type of ideal called a radical ideal will play an important role. To help motivate its definition, think of some more examples where V fails to be injective. Can you spot a pattern? We’ll return to this next time. 

Now that we’ve got our maps V and I it’s instructive to examine their properties. This will give us a feeling for the basic manipulations of algebraic geometry. No need to read it very thoroughly, just skim it to pick up some of the ideas.

Lemma 2.5 The maps I and V satisfy the following, where J_i ideals and X_i subsets of \mathbb{A}^n:
(1) V(o)=\mathbb{A}^n,\ V(\mathbb{A}^n)=0
(2) V(J_1)\cup V(J_2)=V(J_1\cap J_2)
(3) \bigcap_{\lambda\in\Lambda}V(J_{\lambda}=V(\sum_{\lambda\in\Lambda}J_{\lambda})
(4) J_1\subset J_2 \Rightarrow V(J_2) \subset V(J_1)
(5) X_1\subset X_2 \Rightarrow I(X_2)\subset I(X_1)
(6) J_1 \subset I(V(J_1))
(7) X_1 \subset V(I(X_1)) with equality iff X_1 is an affine variety

Proof We prove each in turn.
(1) Trivial.
(2) We first prove “\subset“. Let q\in V(J_1)\cup V(J_2). Wlog assume q \in V(J_1). Then f(q)=0 \ \forall f \in J_1. So certainly f(q)=0 \ \forall f\in J_1\cap J_2, which is what we needed to prove. Now we show “\supset“. Let q\not\in {V(J_1)\cup V(J_2)}. Then q \not\in V(J_1) and q \not\in V(J_2). So there exists f \in J_1, \ g\in J_2 s.t. f(q) \neq 0,\ g(q)\neq 0. Hence fg(q)\neq 0. But fg\in J_1\cap J_2 s0 q \not\in {V(J_1\cap J_2)}.
(3) “\subset” is trivial. For “\supset” note that 0 \in J_{\lambda}\ \forall \lambda, and then it’s trivial.
(4) Trivial.
(5) Trivial.
(6) If p \in J_1 then p(q)=0\ \forall q \in V(J_1) by definition, so p \in I(V(J_1)).
(7) The relation X_1 \subset V(I(X_1)) follows from definitions exactly as (6) did. For the “if” statement, suppose X_1=V(J_1), some ideal J_1. Then by (5) J_1 \subset I(V(J_1)) so by (4) V(I(X_1)=V(I(V(J_1)) \subset V(J_1)=X_1. Conversely, suppose V(I(X_1)= X_1. Then X_1 is the zero set of I(X_1) so an affine variety by definition. \blacksquare

That was rather a lot of tedious set up! If you’re starting to get weary with this formalism, I can’t blame you. You may be losing sight of the purpose of all of this. What are these maps V and I and why do we care how they behave? A fair question indeed.

The answer is simple. Our V,\ I bijections will give us a dictionary between algebra and geometry. With minimal effort we can translate problems into an easier language. In particular, we’ll be allowed to use a generous dose of algebra to sweeten the geometric cocktail! You’ll have to wait until next time to see that in all its glory.

Finally, how does topology fit into all of this? Well, Lemma 2.5 (1)-(3) should give you an inkling. Indeed it instantly shows that the following definition makes sense.

Definition 2.6 We define the Zariski topology on \mathbb{A}^n by taking as closed sets all the affine varieties.

In some sense this is the natural topology on \mathbb{A}^n when we are concerned with solving equations. Letting k=\mathbb{C} we can make some comparisons with the usual Euclidean topology.

First note that since every affine variety is closed in the Euclidean topology, every Zariski closed set is Euclidean closed. However we saw in the last post that not all Euclidean closed sets are affine varieties. In fact there are many more Euclidean closed sets than Zariski ones. We say that the Euclidean topology is finer than the Zariski topology. Indeed the Euclidean topology has open balls of arbitrarily small radius. The general Zariski open set is somehow very large, since it’s the complement of a line or surface in \mathbb{A}^n.

Next time we’ll prove that for algebraically closed k every Zariski open set is dense in the Zariski topology, and hence (if k =\mathbb{C}) in the Euclidean topology. In particular, no nonempty Zariski open set is bounded in the Euclidean topology. Hence we immediately see that the intersection of two nonempty Zariski open sets of \mathbb{A}^n is never empty. This important observation tells us the the Zariski topology is not Hausdorff. We really are working with a very strange topological space!

And how is this useful? You know what I am going to say. It gives us yet another perspective on the world of affine varieties! Rather than just viewing them as geometrical objects in abstract \mathbb{A}^n we can imagine them as a fundamental world structure. We’ll now be able to use the tools of topology to help us learn things about geometry. And there’s the slice of lemon to garnish the perfect cocktail.

I leave you with this enlightening question I recently stumbled upon. Both the question, and the proposed solutions struck me as extremely elegant.