Friday, July 18, 2014

Christoffel symbols



Gauge theory has its roots in general relativity, and we have to start there if we want to properly understand it. (Also general relativity will become our friend and we will come back to it to compare ideas from gauge theory.) An excellent book to learn general relativity is Gravitation and I was very fortunate to learn it as a graduate student directly from Professor Misner. The book's famous saying is: "spacetime tells matter how to move; matter tells spacetime how to curve".

What does it mean spacetime curves? Here I recall an explanation from Gravitation which made a strong impression on me at the time. If you throw a ball you see its curved trajectory as it is attracted by Earth. Similarly, a bullet's trajectory is also curved down, but by a very tiny amount. Why is this huge discrepancy in the amount of curvature in each case? Naively one may say that because of the speed, but how about the equivalence principle? The mass should make no difference in what it basically amounts to a free fall motion where the gravitational effects vanish along the trajectory in an "Einstein's elevator". Therefore one would expect that the curvature is the same!!! And it is the same! Only if one thinks in 3+1 spacetime dimensions, not in 3 space dimensions. Here is the picture from page 33 of the book which explains it all:




Space-time, or in general any Riemannian space is characterized by a metric tensor. When the space is curved, the ordinary derivative has to be modified to take into account changes to the local metric. The corrected derivation operation is called covariant derivative. The difference between the ordinary derivative \( \partial \) and the covariant derivative \( D \) is the Christoffel symbol \( \Gamma \):

\( D_\rho f^\alpha = \partial_\rho f^\alpha + \Gamma^{\alpha}_{\rho \sigma} f^\sigma \)

\( \Gamma \) is also called the affine connection because it helps define the parallel transport along geodesic lines which locally are the straight lines in the curved space ("spacetime tells matter how to move"). Along geodesics, the covariant derivative \( D \) is zero. Gravity does not bend light, There is no dynamical explanation of light bending, but a kinematic one. Light always continues to move locally in a straight line, but specetime itself is warped by the presence of mass. The geodesics are affected by mass. If you ever traveled in an airplane, the trajectory from the starting point to the destination appears curved on the TV screens in the plane. The plane always goes straight from point A to point B on the shortest path to save fuel, but the trajectory appears curved because Earth is not flat, but a curved space (a sphere). The apparent trajectory curvature is encoded by the Christoffel symbols. 

So what is then the covariant derivative of a more complex object like say a tensor \( T^{\alpha \beta}\)?

\( D_{\rho} T^{\alpha \beta} = \partial_\rho T^{\alpha \beta} + \Gamma^{\alpha}_{\rho \sigma} T^{\sigma \beta} + \Gamma^{\beta}_{\rho \sigma} T^{\alpha \sigma} \)

This formula is easily generalized in case of an arbitrary tensor of n indexes: \( \Gamma\) appears n times there. We can apply this to the metric tensor \( g_{\mu \nu}\) and after some algebraic manipulations solve for \( \Gamma \):

\( \Gamma^{\sigma}_{\mu \nu} = \frac{1}{2} g^{\sigma \rho} ( \frac{\partial g_{\rho \mu}}{\partial x^{\nu}} + \frac{\partial g_{\rho \nu}}{\partial x^{\mu}} - \frac{\partial g_{\mu \nu}}{\partial x^{\rho}})\)

The moral of the story is that on Riemann spaces, because the Christoffel symbols depend on the metric, the covariant derivative depends on the metric as well and everything is defined using intrinsic notions. There is only one "fly in the ointment": \( \Gamma \) itself is not a tensor. Why? Because it depends on the partial derivative of the metric tensor with respect of the coordinate system. Change the coordinate system and you get a different \( \Gamma \). What we need is a better mathematical object which changes nicely when the coordinate system is changed. In other words, we need the curvature tensor.

Next time we'll introduce the curvature tensor and we'll talk about a key physical principle: curvature is force.


Friday, July 11, 2014

Short Exact Sequences




We have slowly introduced the tools needed to understand modern physics and we are almost there before seeing them in action. One particular mathematical pattern proves to be very useful and is one of those cases where you need to recognize at first sight: a short exact sequence

We have seen exact sequences before in the homology and cohomology posts where a set of groups \( G_1,...,G_n \) are linked by maps and the image of one map is the kernel of the next map. Then you can only "hop twice" an element through the sequence before you end up in the identity element of the group. The geometric interpretation is that "the boundary of a boundary is zero".

So what happens when the exact sequence is short? The best way to get this is to work the problem from the other end. Suppose we have two sets A and C (I am skipping B on purpose). We can then construct the Cartesian product AxC and for each element from AxC we can understand A as an equivalence class. Then C ~ AxC/A and we can introduce a short exact  sequence: 

\( 0 \rightarrow A \rightarrow (A \times C) \rightarrow C \rightarrow 0\)

Let's first clarify what is with the zeros at each end. The first zero means that the map between A and AxC is injective (one-to-one). This is because the first zero can only be mapped to the zero element of A and that in turn must be the only element which maps to the zero element of AxC. Similarly the last zero element implies that the map between AxC and C is surjective (onto).

Let us now introduce our element B:

\( 0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0\)

Here is the problem: given A and C, what can we say about B?

If A, B, C are groups, here is an example:

  • A = Z (integers)
  • B = Z (integers)
  • C = {0, 1} = Z/2Z (equivalence classes of odd and even numbers, or Z modulo even numbers)
where

  • The first map is the multiplication by 2
  • The second map is the reduction modulo 2: for any \( m \in Z, m=2k+p\) we take k=0. 
So is C ~ B/A:  Z/2Z = {0,1} ~ Z/Z?

Z/Z = {m in Z | m=0} = {0} so the answer is no.

In general knowledge of A and C does not determine B. This is because A and C can interlock in a nontrivial way and B is not necessarily AxC! 

So how about some fancy short exact sequences:

\( 0 \rightarrow S_1 \rightarrow S_3 \rightarrow S_2 \rightarrow 0 \)

\( 0 \rightarrow S_3 \rightarrow S_7 \rightarrow S_4 \rightarrow 0 \)

Those are two of the celebrated Hopf fibrations. I the first case, S3 is a sphere in 4 dimensions and it can be decomposed into an S2 (the ordinary sphere in 3 dimensions) where in each point you have a circle (S1). The picture at the top of this post is a stereographic projection of the rigamarole S3 Hopf fibration because S3 cannot be plotted directly.

The real interesting use of short exact sequences however occurs in physics in gauge theory in fiber bundles.

In this case the short exact sequence pattern is:

\( 0 \rightarrow F (fiber) \rightarrow E (total space) \rightarrow B (base) \rightarrow 0\)

In particular Connes' non-commutative geometry model of the Standard Model is best understood if you work out first a simpler unphysical proposal which gives rise to a short exact sequence. We'll get there...



Friday, July 4, 2014

Exact, Coexact and Harmonic (Hodge Theory)



Happy 4th of July


Today, in celebration of Independence day we'll have some mathematical fireworks :)

It is customary to learned in school about the dot product and the cross product. The dot product comes from projecting one vector onto the other, while the cross product creates a new (pseudo) vector out of two other vectors. The cross product is basically a historical accident which got accepted on due to its practical convenience but a better concept is the exterior product. Even better we can understand all of this in the framework of Clifford algebras.


Here is how it goes. We’ll work out the usual 3D space for convenience. Start with the 3 x,y,z unit vectors and call them: \( e_1, e_2, e_3\). Then introduce 2 practical rules:
  • \( e_1 e_1 = e_2 e_2 = e_3 e_3 = 1 \)
  • \( e_i e_j = - e_j e_i \) when \( i \ne j \)
Think of the unit vectors as matrices which collapse to the identity when multiplied by themselves, and anti-commutes.

Then you can have the following basis in general:
  • scalar: \( 1 \)
  • vectors: \( e_1, e_2, e_3 \)
  • bivectors: \( e_1 e_2, e_2 e_3, e_3 e_1 \)
  • trivector (pseudo scalar): \( e_1 e_2 e_3 = I \)
For two vectors \( A, B\), with \( A = a_1 e_1 + a_2 e_2 + a_3 e_3\) and \( B = b_1 e_1 + b_2 e_2 +  b_3 e_3\) the dot product is:

\(A\cdot B = \frac{1}{2}(AB + BA)\)

and the exterior product is:

\( A \wedge B = \frac{1}{2}(AB - BA) \)

and in general for two vectors:

\( A B = A \cdot B + A \wedge B \)

Here is what we can always do: given a scalar, vector, bivector, or trivector, we can multiply with \( I = e_1 e_2 e_3\) and this defines the Hodge dual \(A \rightarrow \star A \) 

So for example Hodge duality maps bivectors (which are oriented areas to preuso-vectors (the cross product vector orthogonal to the area):

\( A \wedge B = I (A \times B) \)



The Hodge dual exists not only for vectors and bivectors but for differential forms as well:

\( \star dx = dy \wedge dz, \star dy = dz \wedge dx, \star dz = dx \wedge dy \)

The unit volume is: \( vol = I = \star 1 = dx \wedge dy \wedge dz \)

and Hodge defined an inner product of any two p-forms \( \alpha , \beta \) as follows:

\( (\alpha , \beta) = \int <\alpha , \beta > \star (1) = \int \alpha \wedge \star \beta \)

last, Hodge introduces a codifferential \( \delta = {(-1)}^{n(p+1) + 1}\star d \star\)

and proved the Hodge decomposition theorem for any form \( \omega \) :

\( \omega (any form) = d \alpha (exact) + \delta \beta (coexact) + \gamma (harmonic) \)

where \( \Delta \gamma = 0\) Here \( \Delta = d \delta + \delta d\) is the Hodge Laplacian. FIREWORKS PLEASE!!!

Now here is some physics: Maxwell's equations:

Let \( A = A_\mu d x^\mu \) be the electromagnetic four potential. The electromagnetic field 2-form \(F \) is: \( F = dA\)

\( F = \frac{1}{2} F_{\mu \nu}dx^\mu dx^\nu \) with \( F_{\mu \nu}= \partial_\nu A_\mu - \partial_\mu A_\nu \)

Then Maxwell's equations are:

\( dF = 0, d \star F = \star J \)

and the electromagnetic Lagrangian is: \( L = \frac{1}{2} (F, F)\)

So why are we looking at this compact formalism for Maxwell's equations? Because electromagnetism is only one of the 4 fundamental forces in the universe: gravity, weak force, electromagnetism, strong force, and the weak and strong forces are described by Yang-Mills gauge theory which is a generalization of Maxwell's theory. Without a compact notation and a clear geometrical meaning, we have no hope of understanding Yang-Mills theory and we will be stuck forever in the La La land of using cross products, gradients, divergences, and equations in components.

    Saturday, June 28, 2014

    De Rham Theory (part 2)


    Intuitive cohomology


    Continuing from last time we will now introduce the duality between forms and chains, or the Stokes theorem which is of a fundamental importance in math and physics. Recalling that represents the boundary, and d is the exterior derivative, George Stokes



    proved the following identity:



    We can introduce an inner product: between a cycle Ω and a cocycle ω: < Ω , ω > as the integral of ω over the domain Ω.

    Stokes theorem then simply states:  < Ω , ω > = < Ω , dω > Why do we state it like this? Because the boundary of a boundary is zero: ∂∂ = 0 implies that dd = 0 as follows:

    0 = < ∂∂Ω , ω > = < Ω , dω > = < Ω , dd ω >

    Recall that for boundaries we have an exact sequence (which is also called a chain complex). For the usual 3D space this means:



    Then we have a De Rham co-chain complex (in co-chains the arrows point in the reversed direction):



    and from dd = 0 we recognize the usual identities:

    ·        Gradient of a curl is zero
    ·        Curl of a divergence is zero.

    Finally we arrive at De Rham Theorem.

    From chain complexes we extract the Ker/Image homology group: Hp = Zp/Bp
    From cochain complexes we the Ker/Image cohomology group: Hp DR= (α| d α = 0)/( α| d α = β)

    De Rham theorem tells us that the two groups are isomorphic: Hp  ~ Hp DR

    This means that we can explore the topological properties of the space by looking at the solutions of differential equation on that space.

    For example, an electric charge generates an electric field around it. The spatial distribution of the electric field can tell us the location of the charge. That is why the second cohomology group H2 DR can be interpreted as an electric charge! In general cohomological classes have a physical interpretation.


    Next time we will venture into the related wonderful world of Hodge theory.

    Saturday, June 21, 2014

    De Rham Theory


    Intuitive cohomology


    Now we can marry the two lines of argument and arrive at one of the most beautiful and useful advanced mathematical area, de Rham cohomology. I cannot cover this in only one post, and there will be some back to physics posts (one or more guest posts) to prevent the math topics to become too dry. The end goal of the math series is to be able to talk about Yang-Mills theory and the Standard Model, so there is light (physics) at the end of the tunnel.

    Let us start gentle into the topic and consider the real 3 dimensional space. Let us talk about differential forms (the stuff that you pull back). We have 0-forms, 1-forms, 2-forms, and 3-forms. On R3 there are no 4 or higher forms? Why? Read on…

    What 0-forms might be? They are simply the usual functions. Let’s call them f.
    Take the differential of a function:

    df =∂f/∂x dx + ∂f/∂y dy + ∂f/∂z dz

    and you get the gradient which is a 1-form. 1-forms are isomorphic with vector fields X:

    f1 dx + f2 dy + f3 dz    ~  X = (f1, f2 , f3)

    Take the differential of 1-form and you get 2-forms (the curl):

    d ( f1 dx + f2 dy + f3 dz ) =
    +(∂f3/∂y - ∂f2/∂z) dy dz
    – (∂f1/∂z - ∂f3/∂x) dx dz
    + (∂f2/∂x - ∂f1/∂y) dx dy

    Then take the differential of 2-forms and you get a 3-form (the divergence)

    d ( f1 dy dz - f2 dx dz + f3 dx dy ) = (∂f1/∂x + ∂f2/∂y + ∂f3/∂z)  dx dy dz

    The gradient, curl, and divergence are the bread and butter of Maxwell’s equations in college.

    When working with differentials remember 2 rules:

    • (dx dx) = 0
    • dx dy = - dydx

    Now the key idea is the d d = 0 and we are getting somewhere interesting. We will be able to study topology by investigating partial differentials equations. Mathematically this is very surprising because it reveals a bridge between two very different domains. In topology you deal with accumulation points, closed and open sets, while in differential equations you encounter say Schrodinger equation. What can Schrodinger equation tell you about open sets?

    Physically on the other hand it is not surprising at all!!! You know that a violin sounds like a violin, and a drum like a drum. This is because the solution to the wave equation in a cavity depends on the shape of the cavity. By the way, this is why I like physics: it provides an easy context for intractable mathematical abstractions.

    So from above we have:

    • d(0-forms) = gradient
    • d(1-forms) = curl
    • d(2-forms) = divergence

    What is the curl of a gradient? What is the divergence of a curl? They are zero because dd = 0 (recall how long it took to prove those things in college?)

    Now suppose we have a vector field v


    and we ask if it is a gradient of some potential:

    v= grad (A)

    We know that curl (grad (A)) = 0 and locally this is always obeyed, but not globally. If we have a loop (recall homotopy) the integral of v along the loop depends only on the homotopy class and the following vector space:

    v for which curl (v) = 0 BUT v is not grad (A) for some A.

    is the same as the homotopy class!

    In general we can define the following p-th de Rham cohomology (this is a vector space):

                          α | d α = 0
    HpdR = --------------------------------
                           α | α = dβ

    where α is a p-form and β is a (p-1)-form.


    In other words: closed forms which are not exact. This is the same Ker/Image pattern we encountered in homology. To be continued…

    Friday, June 13, 2014

    Tangent vectors, differentials, pushforwards, and pullbacks


    Intuitive cohomology


    Continuing the discussion, we now consider a new way of looking at geometrical spaces. This was started by the sailors who needed maps to guide on their perilous journeys. What people discovered is that they cannot draw a map which was free of distortions: either the distance ratios or the angles (and sometimes both) were not preserved. This is because the Earth is spherical (hence curved) and the map is a flat two dimensional surface. From this mathematicians abstracted the idea of a manifold. A manifold is a geometric shape for which we can locally draw a map. Several maps may be needed to completely cover the space (in the Earth’s case one needs a minimum of two maps), and the collection of maps is naturally called an atlas. An atlas must obey a natural requirement: whenever two maps describe a common area, they have to be compatible.

    Now at any point in the manifold we can attach a tangent space. How do we represent vectors in this plane?




    Here is the first subtle point. The tangent vectors should not be represented by arrows in a plane as in the misleading picture above because this is an embedded picture of the space in a higher dimensional space. As such it can depend on the embedding. What we seek is an intrinsic definition which makes no reference to any embedding. The basis for the tangent plane is given by a linear independent set of partial derivatives along curves passing though the point where the tangent place in defined. Tangent vectors are partial derivatives along some direction!!!

    Next consider a map φ between two manifolds. If point x in M is mapped to point y in N, the map of the two tangent planes at x and y is the differential df. Why? Because a tangent vector is a sum of partial differentials multiplied with some weights, and the differential is the Jacobian which performs a change of coordinates in multi-variable calculus.

    When you first learn differential topology it is customary to write all this in components and build an intuition by repetitive elementary problem solving. However, if you do not pursue the subject further enough you risk of not seeing the forest because of the horrible complicated trees. What we are after is an intuition in coordinate free language. In differential geometry this means we need to understand the key ideas of pushworward and pullback.

    The pushforward is easy to understand. Think of it as a motion picture: a movie on TV is a map between the manifold of where the movie is recorded to the manifold of your screen. Movement on the director’s set is transferred to movement on your screen. In other words, tangent vectors from the domain manifold define tangent vectors in the range manifold.

    The pullback is a bit harder to grasp in an intuitive way. A pullback returns differentials from N to M. Differentials come in many forms. One particular important case is when the manifold N is simply the real number line. In this case the differential is called a 1-form. Think of it as a machine which eats a vector and spits out a number.  The book Gravitation by Misner, Thorne, and Wheeler has this explained the best. If a vector is pictorially represented as an arrow, a 1-form is represented as an infinite set of parallel planes. The distance between the parallel planes is inverse proportional with the 1-form magnitude, while the direction perpendicular with the planes gives the 1-form a direction. A scalar product between a vector and a 1-form corresponds to how many parallel planes the vector pierces. Misner, Thorn, and Wheeler call this “how many bongs of the bell”. Of course if the vector is parallel with the planes (orthogonal with the 1-form) there are no bongs and the scalar product is zero.

    Now back to pullbacks. Imagine at each point on N a 1-form (a set of parallel planes). Can we construct a 1-form on M? Here is how you do it: take a vector from M, push it forward to N, pierce the 1-form and get the number of piercing. Construct on M a 1-form which when pierced by the original vector gives the same “number of bongs”.

    Another way to help visualize pullbacks is by Dirac’s electrons and holes idea. Have a hole in the destination manifold N. We pull back the hole from N to M if we push forward an electron from M to N to fill the hole in N and in the process we created a hole in M.


    To summarize: tangent vectors are partial derivatives, a differential is a map between tangent planes, we pushforward vectors and we pullback differentials. Next time we’ll marry algebraic with differential topology and arrive at the wonderful theory of De Rham cohomology. 

    Friday, June 6, 2014

    What are Betti numbers?


    Intuitive Homology


    We can now wrap the introductory discussion on algebraic topology and build all the key results needed before we will open a new front this time on differential topology. Some mathematical fireworks will follow when we will marry algebraic and differential topology in the beautiful theory of de Rham cohmology.

    We will continue the discussion from last time about holes and introduce three essential concepts: chains, cycles, and boundaries.

    Recall the idea of a simplex. Can we add or subtract two simplexes of order p? The answer is yes and this can be visualized as gluing the simplexes together to form a larger geometrical shape.

    Now this can be generalized and we define a p-chain as a formal sum of p-simplexes. This is important because the set of p-chains form a group: Cp

    In Cp we can define two subgroups:
    • Zp whose elements are p-chains with boundary equal with zero.
    • Bp whose elements are p-chains which are the boundary of (p+1)-chains.

    The groups Zp and Bp may be infinite dimensional, but their quotient group Hp  = Zp /Bp is in many cases finite dimensional and its group dimension is called the Betti number.

    Zp defines a cycle and Bp defines a boundary. So what does exactly Hp corresponding to? Recall the properties of a hole:
    1. It has a boundary
    2. Is not the boundary of anything else



    Q: how many holes does a torus have?
    A: two

     (1) shows that a hole corresponds to an element of Zp  because one can walk in a closed loop around the hole and return to the starting point. From (2) we see that a hole cannot be an element of Bp because a hole is not a boundary of a filled region. Hence a hole corresponds to an element of Zp but not Bp meaning it belongs to the quotient group Hp (and the groups are named appropriately: C for chains, Z for zero, B for boundary, H for hole, or Homology).

    We also see that the boundary operator ∂ is a map between Cp+1 and Cp In fact this map can be extended indefinitely for a particular space (with the understanding that at some point the groups may become trivial):

                      ∂            ∂        ∂         
    …→ Cp+2 →  Cp+1 Cp Cp-1

    When the image of a map becomes the kernel of the next map such a construction is called an exact sequence. Because ∂∂ = 0 (the boundary of a boundary is zero) in homology we have an exact sequence and therefore Hp gives rise to the “Ker modulo Image” pattern.


    A similar pattern will be encountered in differential topology because a double differential is zero as well: dd=0. However in there the arrows will be pointing the other way and instead of homology we will have co-homology. The Betty numbers will be the same because the holes of a space do not change if we discuss chains or integration on the space. Roughly speaking cohomology corresponds to integration on a space, and homolog corresponds to the frontier of that integration.