While I was working on this project, I have also written quite a bit of code (mostly, in Haskell) to help me with my calculations. See the link if you are interested.
Reading free isn’t freedom, one cannot help but think how come some businesses (like Google and Facebook) never charges their “users” but still be one of the richest. I wonder who is the user here.
The main purpose of this post is to prove the polar decomposition theorem for invertible matrices. As an application, we extract some information about the topology of , namely that . Along the way, we recall a few facts and also define real powers of a positive definite Hermitian matrix. We assume that the matrices below are non singular, even though some of the results are true without this assumption. We first prove that every Hermitian matrix has only real eigenvalues. Notice that with respect to the standard Hermitian inner product for a column vector . In fact, more generally, .
Proposition 1 Let be a Hermitian matrix i.e. , then has only real eigenvalues.
Proof: Over complex numbers every non constant polynomial of degree has solutions. Thus, it is clear that has eigenvalues. Let be an eigenvalue of with an eigenvector . That is Taking conjugate transpose of both sides, we get
Since , and we get which is only possible if is real.
Recall that a matrix is called positive definite if for all . Clearly, such a matrix must be nonsingular.
Proposition 2 If is a positive definite matrix, then it has only positive real eigenvalues.
Proof: As above let be an eigenvalue of with an eigenvector . Then,
Thus, must be positive as well.
If is Hermitian, it has an eigenspace decomposition. Here is the sketch of the proof. We apply induction on the dimension . It is clear for . Now, we may assume . Let and be as above. We consider the one dimensional space generated by and the complementary space . By definition, for , , or equivalently . Thus,
or in a more familiar form . This means that preserves which has dimension . So, by induction, has an eigenspace decomposition and we are basically done. The reason that this argument is not very precise is that in this post, we are using a concrete definition of being Hermitian. So, we also have to argue that somehow the matrix is still Hermitian when restricted to . Of course, using the abstract definition, this is trivial. In fact, it is pretty easy to translate every argument presented here to abstract one.
Lemma 3 If is Hermitian, then it is diagonalizable by a unitary matrix.
Proof: Since has an eigenspace decomposition, we can choose a basis of consisting of eigenvectors only. Furthermore, we may choose those vectors to be unit. Consider the matrix that takes the standard basis to this eigenbasis. Then, it is clear that is a diagonal matrix. It is also clear that .
Next, we prove the polar decomposition for invertible matrices. In this proof, we also define the square root of a matrix.
Theorem 4 Given an invertible matrix , there is a positive definite Hermitian matrix and a unitary matrix such that .
Proof: Let . Clearly, i.e. is Hermitian. Also, for nonzero , is nonzero, thus
So, is also positive definite. By the above lemma, as is Hermitian, there is a unitary matrix which diagonalizes i.e. . Since is unitary, and hence, . Also, since is positive definite, all the eigenvalues of and hence of are positive, by the above proposition. So, we define to be where is defined by taking square root of each entry on the diagonal. In fact, using this idea, we can define any power of by . Note that is also diagonal with positive diagonal entries. Hence, in particular, it is Hermitian. Clearly, a diagonal matrix with positive diagonal entries is positive definite. So, is positive definite.
We set . It is easy to check that is positive definite.
as is positive definite.
Finally, we let . Of course, here is invertible because its a product of nonsingular matrices. Now, we just need to check that .
which was to be shown.
Now, we will apply our knowledge to understand the topology of . Given , it is clear from our proof that we can choose positive definite Hermitian part so that . Hence, , in other words, is an element of . Again, in our proof, we have explained that in fact you may take any power of a positive definite Hermitian matrix. So we can define a path of matrices by . We see that and . This defines a deformation retract of onto the . It is easy to see that the space of positive definite Hermitian matrices of determinant 1 is homeomorphic to . More concretely, to write down any such matrix, we need ; . Also, we set . Then,
is positive definite Hermitian of determinant 1.
It is also not very hard to check that only the identity matrix is the only matrix that lies in which is also positive definite Hermitian of determinant 1. Thus, . We leave it as an exercise to prove that .
In this post, we briefly introduce the Lie group , -structures on a manifold and a -manifold. Let us denote the three form on by . We set . This three form is non-degenerate in the sense that whenever we have two linearly independent vectors in , we can find a third vector such that the evaluation of on these vectors is non-zero. We define . One may prove that is a -dimensional Lie subgroup of .
Let us give a different descriptions of . So, it does not look completely arbitrary. is the highest dimension that one may define a cross product. After we identify with the octonions equiped with some octonion product, for any two imaginary octonions we define the cross product to be
Then, we may define the -form on by where the inner product is the standard inner product. Of course, there is a choice on the octonion product and hence, may be different than the one we explicitly wrote above. However, we show that they are equivalent using the right octonion product. To show they are equivalent; first, we prove that is indeed a -form and then, evaluate it on the basis elements to see how the octonion product should be defined.
Using , we obtain and thus, the above definition is equivalent to
To prove that is alternating, it is enough to prove and as we may replace by to get the desired equalities. However, also note that . Therefore, the first two equalities are enough. It is clear that . Hence, we have the first equality. Furthermore,
Thus, we have showed that is alternating.
Our next goal is to define the octonion product. Clearly, from the explicit definition, we want . In other words, . So, a natural choice for the product is . Similarly, we can choose , , , , and . Of course, as we are describing octonion multiplication, we should also define the multiplication with the th generator but it is the generator of part. So, it is just the trivial multiplication i.e. the multiplication coming from the vector space structure. We do not show that this indeed defines an octonion product.
Next, we need to show that they are equal and to do that, it is enough to evaluate on the basis elements. It is an easy computation which we omit.
Note that this definition makes an earlier claim more plausible, namely that is non-degenerate. Because so, we only need to show that is non-zero for linearly independent and . However, that is a built-in property for a cross-product.
A -structure on a manifold can be defined as a subbundle of the frame bundle of whose fibers are isomorphic to . However, there is an equivalent, more convenient definition. In fact, this definition will follow the scheme of the previous post. More explicitly, since fixes on , we may pull it back to each space to have a three form on the manifold and similarly, if we have such a three form on the manifold, then we may find a subbundle of the frame bundle whose fibers are . So, having a three form on such that for any point , and can be identified by an isomorphism between and , means that we can find a -structure on . By an abuse of notation, we call a manifold with a -structure. Furthermore, since is a subgroup of , it also fixes the standard metric and orientation on giving rise to a Riemannian metric and orientation on the manifold. This immediately implies a non-orientable manifold does not admit a -structure.
Next, we introduce -manifolds. Given a manifold with a -structure and the induced metric , let be the Levi Civita connection on . If , we call a torsion free -structure. A manifold with a torsion free -structure is called a -manifold. In fact, there are a number of ways to define -manifolds, as we can see in the following proposition.
Proposition 1 Let be a -structure on with the induced metric and the Levi Civita connection . Then, the following are equivalent:
- and .
If any one of the conditions of the proposition holds (and hence, all), we call a -manifold. The first example of a metric with holonomy is given by Bryant. The metric in his example is incomplete. Later, Bryant and Salamon constructed complete metrics with holonomy on non-compact manifolds. Then, Joyce constructed complete examples on compact manifolds.
Let be a smooth -manifold and be a point in . Consider the set of all linear isomorphisms between the tangent space at and . Note that there is a natural left action of on . Since this action may be seen as a function composition, we denote the action by . Though, we will quite often drop the notation altogether hoping that it is clear. The disjoint union is called the frame bundle of . (Of course, we need to have more conditions on but we will not go into details in this post.) The action of on induces a natural action on .
It is easy to define a bijection between any fiber of and . Using the bijection, we may define a group multiplication on which will, in turn, make the fiber isomorphic to , trivially. In general, this bijection is not canonical as we will see below. First, we fix an isomorphism in . Then, we send to . Clearly, this map is injective and is sent to the identity matrix under this identification. Also, for any , and i.e. the identification is onto. So, we have a bijection. As we can see, once an identity element is fixed, the fiber becomes a group isomorphic to . In other words, is a -torsor.
Let be a Lie subgroup of and be a subbundle of whose fibers (which we still denote by ) are isomorphic to in the above sense. Then, is called a -structure on . Clearly, the frame bundle is a -structure on .
Next, we discuss two examples of proper subbundles inducing various structures on . In our first example, we consider to be the orthogonal group . Recall that the standard euclidean metric on is fixed by . In other words, for any , . We can use this property together with to define a Riemannian metric on . Let , and define the metric as the pullback . We need to show that this definition is independent of the choice of . Let , then as is an -structure. Hence, . So, we can choose any isomorphism in the fiber in order to define . So, we see that an -structure gives us a Riemannian metric. Next, we will go the other way around i.e. given a Riemannian metric , we construct an -structure on . Each tangent space is equiped with an inner product and we consider equiped with the standard inner product. We define a fiber of to be the set of linear isometries between and . Next, we need to check that a fiber is isomorphic to . Again, first, we fix an isomorphism . Then, given another isometry from to , is an isometry from to itself i.e. . Also, for , is an isometry from to and . Hence, as above, we have an isomorphism. It is easy to see that this correspondence between -structures and Riemannian metrics is one to one.
This examle can be generalized easily. Any structure on which is fixed by a Lie subgroup of can be carried to a manifold which admits a -structure. In fact, our second example will be of this type, again. We consider the correspondence between an almost complex structure and a -structure where . Before we discuss the correspondence, let us clearify a few things. We view as a subgroup of using the monomorphism
Let denote the action of on . In other words, where denotes the identity matrix. Or using, the monomorphism defined above
in matrix block form, as a real matrix. Of course, for any matrix , . Equivalently, we have . On the other hand, let . Then,
Therefore, if and only if and . Hence, we see that a real matrix can be identified with a complex matrix if and only if .
Now, we go back to the correspondence between a -structure and an almost complex structure. First, let us assume that we have a -structure and construct an almost complex structure . Let . Then, we define by . Next, we show that is well defined. Let . Since , we have . Hence,
Moreover, clearly, we have . Thus, we have constructed an almost complex structure.
Next, we go in the other direction. Given a complex structure , we form a subbundle of which consists of linear isomorphisms that satisfy . Take a basis of of the form and, then define with in the position and with in the position. It is easy to verify that . Hence, the subbundle is non-empty. Next, we want to show that the fibers are isomorphic to . Let be another isomorphism satisfying . Then, . Thus, , or equivalently, . Therefore, by above remarks. Also, given , that is is also an element of . So, the fibers are isomorphic to .
Our third example will be a -structure. However, I want to have a more detailed discussion of -structures before I present it in this context. So, I will include it in a future post.
There are a number of different statements of Gronwall’s inequality. In this post, we will consider only one of them, perhaps the weakest of all.
Proposition 1 Let be a non-negative continuous function on such that there are positive constants and satisfying
for all . Then,
for all .
Proof: Define . Note that, by definition, and is a strictly positive differentiable function. Also, we have . In other words, which means the relative rate of change of is less than . Hence, the growth of is slower than an exponential function with relative rate of change . That is (if you did not like this reasoning, you may integrate both sides of the previous inequality from to ). So, we have the desired result .
Consider a real vector space generated by and . There is an obvious identification of with the complex plane such that and . Define a linear complex structure on by setting and . With the identification mentioned above, since is a complex vector space, can be viewed as a complex vector space, too. Furthermore, the action of can be viewed as multiplication by on but we will see below why this view does not extend further.
Next, we complexify by taking a tensor product with over . We know that (real) dimension of is and it is generated by and . We can view as a complex vector space and, for notational simplicity, write and . Note that over the complex numbers is dimensional and generated by and . However, these are not the “natural” bases to work with as we wil see. Next, we extend (complexify) to get which we will still denote by for notational simplicity. Let and . Now, we see that
This means that is an eigenvector of corresponding to the eigenvalue . Similarly, is an eigenvector corresponding to the eigenvalue . So, the set is an eigenbasis for and it gives us an eigenspace decomposition of . Computing , using this basis, is clearly more convenient and hence, this is a “natural” choice as a basis. Furthermore, from this viewpoint, it is also clear why the action of cannot be viewed as multiplication by any more.
The following one line of Haskell code is a very simple version of the cat.
main = interact id
Disclaimer: this post is opinion based.
A few years ago, I have tried to write a JS code to draw some fractals. However, I was not able to come up with an original and beautiful result. So, I have modified the code a little and made it draw a bunch of lines with random colors (and repeat that every second.) You can view it here.
Back then, I did not see anything particularly nice about it but after a couple of years, when I look at it again, it makes me think about our perception of difficulty. Let me emphasize that it draws exactly the same lines at every run, it is just the colors that are random (unless, of course, I have made an error.) So, in theory, no matter which set of colors you start with, if you diligently trace the lines, you can understand the pattern of the lines. Here is a sample picture:
I do not know about you but the above picture does not give me a lot of hints about its pattern other than a possible point of symmetry. So, I would say that it is hard to recognize a pattern, if it exists at all.
As you saw, it became a lot easier to notice a pattern when you change the way you look at the lines. A quite similar situation happens often in my daily life; when I spend a lot of hours working on a seemingly hard mathematical problem, only to realize that I have been using wrong “colors”. Of course, in general, you do not know which “colors” to start with but if you find something challenging, it is often rewarding to change “colors”.