# Decomposition of Curvature Tensors

It’s been a long time since I last posted something here. Today is a rainny day. It is special to make a few notes.

An interesting fact has been used regularly in Riemannian geometry is that the curvature tensor of a Riemannian manifold splits naturally into three orthogonal components, involving respectively its scalar curvature, trace-less part of its Ricci tensor and its Weyl tensor. This is classical and well known but surprisingly, this does not appear in most textbooks on Riemannian Geometry.

The mystery behind this fact comes from one of the most powerful tools in mathematics, Representation Theory. The key point is that the bundle which the curvature tensor lives is not irreducible under the action of orthogonal group and consequently is decomposed into irreducible components. These gives a canonical decomposition of the curvature tensor. In this post, I am going to show explicitly how it actually works.

Let us get down to the details. For now, we always assume that $(M,g)$ is an $n$-dimensional Riemannian manifold. It is well known that the curvature tensor $\mathrm{Rm}$ is decomposed into three components $\mathrm{Rm} = \frac{\mathrm{Scal}}{2n(n-1)} g \mathbin{\bigcirc\mspace{-15mu}\wedge\mspace{3mu}} g + \frac{1}{n-2} g \mathbin{\bigcirc\mspace{-15mu}\wedge\mspace{3mu}} \stackrel{\circ}{\mathrm{Ric}} + W$

where $\stackrel{\circ}{\mathrm{Ric}} = \mathrm{Ric} -\frac{\mathrm{Scal}}{n} g$ denotes the traceless Ricci tensor, $W$ denotes Weyl tensor and $\mathbin{\bigcirc\mspace{-15mu}\wedge\mspace{3mu}}$ denotes the Kulkarni-Nomizu product.

First, I want to recall some facts in terms of the curvature tensor of a Riemannian manifold. The curvature tensor satisfies the following properties:

1. $\mathrm{Rm}(X,Y,Z,W)=-\mathrm{Rm}(Y,X,Z,W)=\mathrm{Rm}(Z,W,X,Y)$.
2. (The first Bianchi identity) $\mathrm{Rm}(X,Y,Z,W)+\mathrm{Rm}(Y,Z,X,W)+\mathrm{Rm}(Z,X,Y,W)=0, \forall X,Y,Z,W \in \Gamma(TM)$.

The first property means that we can consider the curvature tensor as a section of the subbundle $\mathrm{Sym}^2(\Lambda^2 M)$ of the vector bundle $\bigotimes^4 TM$.

Now, forget about the manifold structure. Let $(E,g)$ be an $n$-dimensional Euclidean vector space. We can understand the general vector space $E$ as the tangent space of the manifold. Denote $\mathrm{Sym}^{2}_{b} (\Lambda^2 E)$ as the subspace of $\mathrm{Sym}^2 (\Lambda^2 E)$ contains all of elements satisfying the first Bianchi identity. A vector in $\mathrm{Sym}^{2}_{b} (\Lambda^2 E)$ is called an algebraic curvature tensor. If we can prove that the vector space of algebraic curvature tensor as a $O(n)$-module can be decomposed into irreducible components, then the decomposition of the curvature tensor follows.

We consider the representation of the general linear group $GL(E)$ on the vector space $T^{(k,l)} E= \bigotimes^k E^{*} \otimes \bigotimes^l E$. The action of $GL(E)$ on $T^{(k,l)} E$ is as follows. For any $L \in GL(E)$, $\omega_1, \ldots, \omega_k \in E^{*}, e_1, \ldots, e_l \in E$, $L(\omega_1 \otimes \ldots \otimes \omega_k \otimes e_1 \otimes \ldots \otimes e_l)=(\omega_1 \circ L^{-1}) \otimes \ldots \otimes (\omega_k \circ L^{-1}) \otimes (L e_1) \otimes \ldots \otimes (L e_l).$

The metric $g$ induces a canonical identification between $E$ and $E^{*}$. If $L$ belongs to the orthogonal group $O(n)$, then by $L^{-1}=L^{T}$, $E$ and $E^{*}$ are isomorphic as $O(n)$-modules. Therefore, we may consider tensor products of $E$ only.

It is easy to check that $O(n)$-module $E$ is irreducible ( this means there is no proper subspace of $E$ invariant under the action of the group $O(n)$). However, it is well known that $\bigotimes^2 E$ is not irreducible. So there might be a decomposition of $\bigotimes^2 E$ into irreducible components. I finish my post here by stating a result that answers this question. I will provide the proof for this result for the next posts.

Proposition 1. The irreducible orthogonal decomposition of the $O(n)$-module $\bigotimes^2 E$ is the following $\bigotimes^2 E= \Lambda^2 E \oplus \mathrm{Sym}_{0}^{2} E \oplus \mathbb{R}g$

where $\mathrm{Sym}_{0}^{2} E$ denotes the space of traceless symmetric 2-tensors.

References:

 Besse, Arthur L. (2008). Einstein Manifolds. Classics in Mathematics. Berlin: Springer. ISBN 978-3-540-74120-6.

 H. Weyl, Classical groups, their invariants and representations, Prineton University Press, Princeton (1946).

 R.S.Hamilton, Three manifolds with positive Ricci curvature, J.Diff.Geom. 17, 255-306 (1982).

# Dissertation

I upload here my undergraduate dissertation that I have just completed lately.

# The talk at Midnight!

That was not an ordinary talk. I guess it would be that cured a relationship. I don’t expect I could go further from that but at least having a relationship is really good for me at the moment. I would love to write in the past tense because the past means memories and the one whom I want to mention in this post is one of the first people who made me have a crush on.

That is a normal girl. She is my classmate in high school. I realized that I had a crush on her at the end of our high school studies. My feelings for her were as pure as those of other schoolboys at that time. I am not sure whether she was my first love or not but actually she gave me special emotions when I faced or stood by her. No matter how anyone judges her and whatever she wants, in my eyes, she has been always the girl with pure thoughts. Unfortunately, we haven’t been talking for a long time. I don’t remember exactly what happened at that time but I guess it came from my immaturity. After that, I don’t know how to recover our relationship.

Last night, I incidentally saw a story that she posted on her Instagram. It was so surprising that the post is about one of my favorite movies. I thought that was a good chance to restart but I had a little hesitation. I asked myself how she would think of that and how she would respond. After a couple of seconds, I decided to text her. I seemed to be ready to receive a bunch of bad words from her. Fortunately, there was nothing terrible happening, I felt like she also wants to maintain a good relationship with me. I was able to realize that through how she talked to me. That ‘s so good and beyond my imagination. Then, we continued to discuss the movies and I made some jokes in the hope that it may remind her about what we have experienced before. I am not sure how she thinks but that was a really good night.

I have no idea about how our relationship would become but to be able to maintain it is still a good thing because we used to be friends.

# An amazing trick

It ‘s been 2 weeks since I arrived in Hong Kong. I still cannot get used to the lifestyle here. Everything seems to be strange to me. Here I experience for the first time simultaneously loneliness, pressure, boredom, etc. Today is boring Sunday again. As usual, I get up and go to the library. I think it ‘s the best place that I can come to at the moment. Recently, I have tried to read a well-known Hamilton’s paper that my advisor provided me. It ‘s really difficult paper for those who have just started to study Riemannian Geometry like me. I cannot virtually get anything from the paper these days. Fortunately, during the time I have struggled to understand the paper, I found a great trick that may help us reduce considerably our computation in terms of “tensorial” quantities. Walk on the campus and think about the paper, I thought it ‘s better to note something on my blog.

First of all, I would like to mention the motivation of this post. Suppose $M$ is a Riemannian manifold, we concern about several geometric flows that derive from evolution equations like $\frac{\partial g}{\partial t} = v(t)$

where $g(t)$‘s is a family of Riemannian metric and $v(t)$‘s is a family of some symmetric to tensor. In local coordinate, we may write equations as $\frac{\partial g_{ij}}{\partial t} = v_{ij}$ where $g=g_{ij} \mathrm{d}x^{i} \otimes \mathrm{d}x^{j}$ and $v=v_{ij} \mathrm{d}x^{i} \otimes \mathrm{d}x^{j}$. In several cases, we may be interested in finding the rate of change of Christoffel symbols $\Gamma_{ij}^{k} (t)$ ‘s.

To understand this post, readers need to have some basic background on Riemannian Geometry. First of all, I would like to introduce a notion called geodesic normal coordinates. The following result states its existence.
Proposition Let $(M,g)$ be a Riemannian manifold. Then at every $p \in M$, there exists a local coordinate ${u_i}$ at p such that all of the following hold:
1. $g_{ij}:=\delta_{ij}$ for any $i,j$ at the point $p$ ;
2. $\Gamma_{ij}^{k}=0$ for any $i,j,k$ at $p$ ;
3. $\frac{\partial g_{ij}}{\partial u^k} =0$ for any $i,j,k$ at $p$
This local coordinate is called the geodesic normal coordinate

Now return to the main problem. We fix a point and a time $(p, t_0)$. Choose the normal coordinate ${u^i}$ at $p$ respect to the metric $g(t_0)$.Recall that Christoffel symbols given by Levi-Civita connection as $\Gamma_{ij}^{k}=\frac{1}{2} g^{kl}\left(\frac{\partial g_{jl}}{\partial u^i}+\frac{\partial g_{il}}{\partial u^j}-\frac{\partial g_{ij}}{\partial u^l} \right)$

differentiating to both sides, we have $\frac{\partial \Gamma_{ij}^{k}}{\partial t} = \frac{1}{2}\left(\frac{\partial g^{hl}}{\partial t} \right) \left( \frac{\partial g_{jl}}{\partial u^i}+\frac{\partial g_{il}}{\partial u^j}-\frac{\partial g_{ij}}{\partial u^l} \right) + \frac{1}{2}g^{kl}\left(\frac{\partial v_{jl}}{\partial u^i}+\frac{\partial v_{il}}{\partial u^j}-\frac{\partial v_{ij}}{\partial u^l} \right)$.

That is where the amazing thing comes out. At the point $p$ with the normal coordinate, we can simplify our expression above by the fact that the term $\frac{1}{2}\left(\frac{\partial g^{hl}}{\partial t} \right) \left( \frac{\partial g_{jl}}{\partial u^i}+\frac{\partial g_{il}}{\partial u^j}-\frac{\partial g_{ij}}{\partial u^l} \right)$ vanishes.

Therefore, $\frac{\partial \Gamma_{ij}^{k}}{\partial t} = \frac{1}{2}g^{kl}\left(\frac{\partial v_{jl}}{\partial u^i}+\frac{\partial v_{il}}{\partial u^j}-\frac{\partial v_{ij}}{\partial u^l} \right)$.

Also, In the normal coordinate, due to the fact that Christoffel symbols vanish, the partial derivative will turn into the covariant derivative by $\partial_{k} v_{ij} = \frac{v_{ij}}{\partial u^k} - v_{lj}\Gamma_{ki}^{l}-v_{il}\Gamma_{kj}^{l}.$

Finally, we get $\frac{\partial \Gamma_{ij}^{k}}{\partial t} = \frac{1}{2}g^{kl}\left(\partial_{i} v_{jl} +\partial_j v_{il}-\partial_l v_{ij} \right)$ at $p$.

Now, we can see that the LHS is a (1,2)-tensor because the difference of Levi-Civita connections is actually a tensor ( readers can easily check it out by using the definition of Levi-Civita connection). Meanwhile, the RHS is also a tensor of the same kind as it is the contraction of other tensors. To sum up, the RHS and LHS now are independent of the choice of coordinate charts. Because the point $p$ vary arbitrarily, we eventually obtain that $\frac{\partial \Gamma_{ij}^{k}}{\partial t} = \frac{1}{2}g^{kl}\left(\partial_{i} v_{jl} +\partial_j v_{il}-\partial_l v_{ij} \right)$ for everywhere.

Similarly, we can also establish variations of more complicated quantities such as (1,3)-, (0,4)-Riemmanian tensors, etc.

If we take $v_{ij}=-2R_{ij}$, we get celebrated Ricci flow, that was proposed by Hamilton and then used by Perelman to solve famous Poincare conjecture. Such kinds of equalities are really important for analyzing the evolution of curvatures and by that way, Poincare conjecture was solved.