## simple update

It’s been already like a year since I posted some maths articles here, and I do not know what I have done since. This winter has been very cold and cold for a long time, and so it is still a bit cold to go outside. I thought when it comes to winter, I could stay in my office all day long, not wanting to get out of the building, and study very hard, but reality was… I did not even want to get out of my house. Also, studying at home is… you know, very distracting.

I have recently started to study a new thing, and if I turn out to be diligent, I would post some things about what I am reading and hopefully make this blog more active. I do not remember much about the things I have posted, so I am not sure if I can finish the 7-sphere thing; however, if I get very enthusiastic, I can figure out what to write and finish it.

I do not know if there is anyone who keeps up with this blog. If there is, thank you! I will see you soon, hopefully! 🙂

## 27 lines on a cubic surface

I recently had a chance to study a little bit about this subject and presented in a ‘graduate seminar’ course. I am attaching the note to prepare for this presentation. There are many ways to prove this interesting fact, but I took the one of the easiest way to prove it, namely, using Complex manifolds theory. Here is the note.

1. What is a Smooth Cubic Surface?

We first define what a smooth cubic surface is. For simplicity, we will work on the complex field ${\mathbb C}$.

Definition 1 An ${n}$-dimensional projective space ${\mathbb P^n}$ is a space consists of lines in ${(n+1)}$-dimensional complex space ${\mathbb C^{n+1}}$ which pass through the origin. In other words,

$\displaystyle \mathbb P^n := \left(\mathbb C^{n+1}-\{0\}\right)/\sim$

where the equivalence is given by

$\displaystyle z=(z_0,\cdots,z_n)\sim (w_0,\cdots,w_n)=w \iff \exists t\in \mathbb C\backslash\{0\} \text{ s.t. } z=tw.$

For a projective space ${\mathbb P^n}$, we use the homogeneous coordinate. That is, the point corresponding to ${z=(z_0,\cdots,z_n)\in \mathbb C^{n+1}-\{0\}}$ is ${[z_0;\cdots;z_n]}$ where the equivalence induces the relation ${[tz_0;\cdots;tz_n]=[z_0;\cdots;z_n]}$ for ${t\neq 0}$.

Remark 1 For ${z=[z_0;\cdots;z_n] \in \mathbb P^n}$, there is ${i}$ that has nonzero ${z_i}$. Then, there is a homeomorphism between ${\{z\in \mathbb P^n| z_i\neq 0\}}$ and ${\mathbb C^n}$ given by ${z\mapsto \frac{1}{z_i}(z_0,\cdots, \hat z_i,\cdots, z_n)}$. Thus, ${\mathbb P^n}$ is covered by ${n+1}$ copies of ${\mathbb C^n}$. Thus, ${\mathbb P^n}$ is a (smooth) manifold.

From now on, we will mostly consider the case when ${n=3}$. That is, ${\mathbb P^3}$. Also, we will denote the coordinate by ${[x;y;z;w]}$ instead of ${[z_0;z_1;z_2;z_3]}$.

Observe that a polynomial function is not well-defined on ${\mathbb P^3}$. However, if the polynomial is homogeneous, i.e. degree of each monomial is all identical, the inverse image of zero(=zero locus) is well-defined.

Definition 2 A smooth cubic surface is a degree 3 hypersurface in ${\mathbb P^3}$ that is smooth. More precisely, a smooth cubic surface is the zero locus of a degree 3 homogeneous polynomial ${f(x,y,z,w)}$ with the non-zero jacobian

$\displaystyle \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}, \frac{\partial f}{\partial w}\right)\neq 0.$

Remark 2 In fact, the more precise definition would be the zero locus of a degree 3 homogeneous polynomial where for each point ${p\in \mathbb P^3}$ such that ${f(p)=0}$, the jacobian at ${p}$ is of rank 1. However, the previous definition suffices in this case because

$\displaystyle 3f=x\frac{\partial f}{\partial x}+y \frac{\partial f}{\partial y} + z \frac{\partial f}{\partial z} + w \frac{\partial f}{\partial w}$

and thus if any point in ${\mathbb P^3}$ has rank 0 jacobian matrix, then it must be in the zero locus.
Additionally, we remark that irreducibility of ${f}$ is guaranteed by the Jacobian criterion. If ${f=gh}$ for some non-constant homogeneous polynomials ${g}$ and ${h}$, then by Bèzout’s theorem, there is ${p\in \mathbb P^3}$ such that ${g(p)=h(p)=0}$. This guarantees that the Jacobian of ${f}$ at ${p}$ is zero.

The simplest example of a smooth cubic surface is when ${f=x^3+y^3+z^3+w^3}$. We call it the Fermat cubic.

2. Cayley-Salmon theorem

Definition 3 By a line in a smooth cubic surface ${S\subset \mathbb P^3}$, we mean a line in the projective space ${\mathbb P^3}$ that is contained entirely in ${S}$.

We are now ready to state the main theorem.

Theorem 4 (Cayley-Salmon) Any smooth cubic surface contains exactly 27 lines.

In case of the Fermat cubic, we can actually formulate all those 27 lines:

Note that a line in the projective space ${\mathbb P^3}$ is given by two hyperplanes(=hypersurface of degree 1). Choose a line ${L}$ in ${\mathbb P^3}$. Up to the coordinate exchange(permutation of ${x,y,z,w}$), if necessary, those two hyperplanes can be given as

$\displaystyle x=az+bw, y=cz+dw.$

Therefore the line ${L}$ is contained in the Fermat cubic ${S}$ if and only if

$\displaystyle a^3+c^3+1=0,\quad b^3+d^3+1=0,\quad a^2b+c^2d=0,\quad ab^2+cd^2=0$

Observe that, if ${abcd\neq 0}$, then from relation 3 and 4, ${(a^2b)^2/(ab^2)=a^3=(-c^2d)^2/(-cd^2)=-c^3}$, thus it contradicts to the first relation. Therefore, at least one of those coefficients is zero. Again, by the coordinate exchange, we may assume that ${a=0}$.

If ${a=0}$, then we get ${c^3=-1}$ and ${d=0, b^3=-1}$. Therefore ${a=d=0}$ and ${b^3=c^3=-1}$. The candidate for ${(b,c)=(-\zeta^k,-\zeta^l)}$ where ${\zeta}$ is the primitive 3rd root of unity and ${0\leq k,l\leq 2}$. That is, in this case, the lines are of the form

$\displaystyle x+w\zeta^k=y+z\zeta^l=0$

and by the permutation, we get also get

$\displaystyle x+z\zeta^k=y+w\zeta^l=0,\quad x+y\zeta^k=z+w\zeta^l=0$

This shows that there are exactly 27 lines on the Fermat cubic.

3. Proof

The tactic for the proof is the following: We will show that the number of lines contained in a smooth cubic surface is invariant as we vary the cubic polynomial that has non-zero jacobian in ${\mathbb P^3}$. From the fact that the Fermat cubic has 27 lines in it, the theorem is proved.

3.1. Preparation for the proof

We introduce spaces that take important roles to prove this theorem.

First, the space that parametrise cubic surfaces in ${\mathbb P^3}$ is ${\mathbb P^{19}}$, because there are ${\begin{pmatrix}3+3\\3\end{pmatrix}=20}$ different degree 3 monomials with 4 variables, and the constant multiplication does not affect the zero locus, which, in fact, is the cubic surface. From now on, we will identify an element of ${\mathbb P^{19}}$ with a cubic surface in ${\mathbb P^3}$ as long as the polynomial is irreducible.

Lemma 5 The subset ${U}$ of ${\mathbb P^{19}}$ that consists of elements corresponding to SMOOTH cubic surfaces forms a Zariski open subset in ${\mathbb P^{19}}$. That is, ${U}$ is an open dense subset and the complement consists of finite union of closed sets of positive codimension.

Second, the space that parametrise lines in ${\mathbb P^3}$ is the Grassmannian ${Gr(2,4)}$, because a line in ${\mathbb P^3}$ corresponds to a 2-dimensional subspace in ${\mathbb C^4}$. An element in ${Gr(2,4)}$ consists of two linearly independent vectors in ${\mathbb C^4}$, and an equivalence relation given by the subspace they span. Therefore, elements of the Grassmannian ${Gr(2,4)}$ are ${2\times 4}$ RREF matrices of rank 2. Another characterisation of the Grassmannian space gives the following property:

Lemma 6 The (complex) Grassmannian space ${Gr(2,4)}$ is homeomorphic to ${U(4)/(U(2)\times U(2))}$, with the obvious quotient action. In particular, ${Gr(2,4)}$ is compact.

Finally, we are ready to describe the space we want. Consider the subspace ${M}$ of the product space ${\mathbb P^{19}\times Gr(2,4)}$ defined by

$\displaystyle M:=\{(X,L)| X\in U, L\subset X\}$

where ${U\subset \mathbb P^{19}}$ is the open subset consisting of smooth cubic surfaces.

3.2. Proof

Proposition 7 The space ${M}$ is locally a zero locus of polynomials in ${U\times Gr(2,4)}$.

Proof: Observe first that with change of coordinate on ${\mathbb P^3}$, we can represent any point ${p=(X,L)\in M}$ with ${L=\{z=w=0\}\subset \mathbb P^3}$. Therefore, the open set around the point ${L\in Gr(2,4)}$ is of the form

$\displaystyle \begin{bmatrix} 1&0&a&b\\ 0&1&c&d \end{bmatrix}$

with the case ${a=b=c=d=0}$ corresponds to ${L}$.

Write ${f_X}$ to denote a cubic polynomial corresponding to a cubic surface ${X}$. Then

$\displaystyle \begin{array}{rl} L\subset X &\iff f_X(s(1,0,a,b)+t(0,1,c,d))=0 \quad \forall s,t\\ &\iff f_X(s,t,as+ct,bs+dt)=0 \quad \forall s,t\\ &\iff \sum_{i=0}^3 s^i t^{3-i}F_{X,i}(a,b,c,d)=0 \quad \forall s,t\\ &\iff F_{X,i}(a,b,c,d)=0 \quad i=0,1,2,3. \end{array}$

This shows that ${M}$ is locally zero locus of ${F_{X,i}}$‘s. $\Box$

Proposition 8 Choose a point ${p\in U}$. Then for the projection ${\pi:M\rightarrow U}$, there is a neighbourhood ${V}$ of ${p}$ in ${U}$ such that for each ${p_0\in \pi^{-1}(p)}$, there is a unique (continuously differentiable) section on ${V}$. That is, each connected component of ${\pi^{-1}(V)}$ is homeomorphic to ${V}$ by the map ${\pi}$ restricted to the connected component.

Proof: To prove this proposition, we will use the implicit function theorem:

Lemma 9 (Implicit Function theorem) Suppose ${f:\mathbb C^{n+m}\rightarrow \mathbb C^m}$ is a continuously differentiable function. Write ${x\in \mathbb C^n}$ and ${y\in \mathbb C^m}$, so that ${(x,y)\in \mathbb C^{n+m}}$. If the Jacobian matrix ${\left( \frac{\partial f_i}{\partial y_j}\right)_{ij}}$ is invertible at ${(a,b)\in \mathbb C^{n+m}}$, then there is a neighbourhood ${\tilde U}$ of ${a\in \mathbb C^n}$ and the unique continuously differentiable function ${g:\tilde U\rightarrow \mathbb C^m}$ s.t ${g(a)=b}$ and ${f(a,g(a))\equiv c}$.

We apply this theorem to ${F=(F_1,F_2,F_3,F_4)}$ where ${F_i(X,\cdot)=F_{X,i}}$, which is defined on the open neighbourhood of ${p=(X,L)}$. That is, if we find out that ${\frac{\partial(F_{X,1},F_{X_2},F_{X,3},F_{X,4})}{\partial (a,b,c,d)}(0,0,0,0)}$ is invertible, then there is an open neighbourhood ${V}$ of ${X\in U}$ and the unique continuously differentiable function ${g}$ such that ${g(X)=L}$ and ${(X,g(X))\in M}$.

Consider

$\displaystyle \frac{\partial (\sum s^i t^{3-i} F_{X,i})}{\partial a}(0,0,0,0)=\frac{\partial f_X(s,t,as+ct,bs+dt)}{\partial a}|_{a=b=c=d=0}=s\cdot \frac{\partial f_X}{\partial z}(s,t,0,0).$

Similarly, \begin{align*} \frac{\partial (\sum s^i t^{3-i} F_{X,i})}{\partial b}(0,0,0,0)&=s\cdot \frac{\partial f_X}{\partial w}(s,t,0,0),
\frac{\partial (\sum s^i t^{3-i} F_{X,i})}{\partial c}(0,0,0,0)&=t\cdot \frac{\partial f_X}{\partial z}(s,t,0,0),
\frac{\partial (\sum s^i t^{3-i} F_{X,i})}{\partial d}(0,0,0,0)&=t\cdot \frac{\partial f_X}{\partial w}(s,t,0,0). \end{align*}

Since ${s^i t^{3-i}}$ formally distinguishes the row of the Jacobian matrix, if this Jacobian matrix is not invertible, we must have

$\displaystyle (s\lambda_1 + t\mu_1 )\frac{\partial f_X}{\partial z}(s,t,0,0)+(s\lambda_2 + t\mu_2)\frac{\partial f_X}{\partial w}(s,t,0,0)=0$

for some ${(\lambda_1,\lambda_2,\mu_1,\mu_2)\in \mathbb C^4-\{0\})}$.

Since ${f_X}$ is a homogeneous polynomial, ${\frac{\partial f_X}{\partial z}(s,t,0,0)}$ and ${\frac{\partial f_X}{\partial w}(s,t,0,0)}$ are both homogeneous polynomial with 2 variables over the algebraically closed field ${\mathbb C}$, thus they are factored into linear terms. Notice that there are at least two zeros for ${\frac{\partial f_X}{\partial z}(s,t,0,0) }$ and therefore to get the above equation for both zeros, we must have a common zero ${p=(p_1,p_2,0,0)\in L}$ so that ${\frac{\partial f_X}{\partial z}(p)=\frac{\partial f_X}{\partial w}(p)=0}$.

Since ${f_X(s,t,0,0)=0}$ for all ${s,t}$, we get ${\frac{\partial f_X}{\partial x}(p)=\frac{\partial f_X}{\partial y}(p)=0}$. Therefore, ${p}$ is a singular point in a smooth cubic surface ${X}$, which is the contradiction. Therefore the Jacobian matrix is invertible and we are done. $\Box$

Proof: For the projection ${\pi: M \rightarrow U}$, fix a point ${X\in U}$. From the propositions above, if the fibre ${\pi^{-1}(X)}$ is of cardinality ${k}$, then there is a neighbourhood of ${X}$ in ${U}$ such that any fibre in the neighbourhood is of cardinality larger of equal to ${k}$. Also, since ${M}$ is locally closed, if ${(X,L)\notin M}$, then there is a neighbourhood of ${(X,L)}$ in ${U\times \mathbb Gr(2,4)}$ such that the neighbourhood is disjoint from ${M}$. Since ${Gr(2,4)}$ is compact, we can find a neighbourhood ${V_X}$ of ${X}$ in ${U}$ so that ${\pi^{-1}(V_X)=\coprod_{i=1}^k V_X}$. That is, the cardinality is locally constant.

Since ${U}$ is connected, and the Fermat’s cubic contains 27 lines, we conclude that all smooth cubic surfaces contains 27 lines. $\Box$

4. Furthermore, and References

There are many other ways to prove this theorem. One another way is purely algebro-geometric. You can find out the result in many reasonably thick algebraic geometry books. There is also a proof using Chern class of the 3-symmetric line bundle of the tautological line bundle over the Grassmannian ${Gr(2,4)}$, and a proof using representation theory – which I do not know well, but it is related to the Weyl group ${E_6}$.

Moreover, there is a theorem saying that all smooth cubic surfaces are 2-dimensional projective spaces ${\mathbb P^2}$ blown up at 6 points.

References for this mini note is the following:

A lecture note of Andreas Gathmann
http://www.mathematik.uni-kl.de/${\sim}$gathmann/class/alggeom-2014/chapter-11.pdf

A lecture note of Ravi Vakil
http://math.stanford.edu/${\sim}$vakil/216blog/FOAGfeb0717public.pdf

AMS Blog on this topic
http://blogs.ams.org/visualinsight/2016/02/15/27-lines-on-a-cubic-surface/

Posted in Maths | 1 Comment

## Exotic 7-sphere 2

**For simplicity, one says Closed manifold’ to denote Compact manifold (without boundary)’ **

It’s been forever since I have written anything on my blog. I am not busy but busy, so completely forgot about this post. Now, here we go.

To talk about the generalised Poincaré conjecture, we can think about which part of the conjecture can be generalised. Obviously, the dimension can be generalised: we may consider whether the conjecture still holds if we consider ${n}$-dimensional manifold for ${n>3}$. However, there are easy counter-examples when ${n}$ is larger than 3: ${S^2\times S^{n-2}}$ is a simply connected ${n}$-dimensional manifold not homeomorphic to ${S^n}$. This seems to be too obvious: this generalisation doesn’t seem to be right (at least, it is not so interesting at all). Maybe there should be some hypothesis that has to vary as the dimension changes. Observe the following corollary derived from Huwewicz theorem and Whitehead Theorem:

Corollary 1 A continuous map ${f:X\rightarrow Y}$ between two simply connected CW-complexes which induces isomorphism on (integer) homology groups is a homotopy equivalence.

Now, let ${M}$ be a simply-connected closed manifold of dimension 3. We claim that the manifold ${M}$ is a homotopy 3-sphere, that is, by definition, homotopy equivalent to the 3-sphere ${S^3}$.

Lemma 2 A simply-connected closed 3-manifold ${M}$ is a 3-sphere.

There are many ways to prove it, but the way I like is the following:

Claim 1 The manifold ${M}$ is orientable.

proof of Claim 1: Suppose the manifold ${M}$ is non-orientable. Then there is a double cover ${\tilde M}$, which is a orientable connected manifold. Let ${\{x_0, x_1\} \subset \tilde M}$ be the fibre of a point ${x\in M}$. Consider a path ${\alpha: [0,1]\rightarrow \tilde M}$ such that ${\alpha(0)=x_0}$ and ${\alpha(1)=x_1}$. This path projects to a loop in ${M}$ based on the point ${x\in M}$. Since ${M}$ is simply connected, the projected loop is homotopic to a constant loop, which cannot happen as the homotopy downstair can be lifted up to an homotopy and a path (with fixed different ends) cannot be homotopic to a constant path. This contradiction says that ${M}$ is necessarily orientable.///

Claim 2 The homology group of ${M}$ is the same as that of ${S^3}$.

proof of Claim 2: Clearly ${H_0(M;{\mathbb Z})=H_3(M;{\mathbb Z})={\mathbb Z}}$ and ${H_1(M;{\mathbb Z})=0}$ by the connectedness, orientability and simply-connectedness. From the universal coefficient theorem, we get ${H^1(M;{\mathbb Z})=0}$ and therefore by Poincaré Duality, we get ${H_2(M;{\mathbb Z})=0}$. Thus, we are done.///

By the (absolute version of) Hurewicz theorem, we have ${\pi_1(M)=\pi_2(M)=0}$ and ${\pi_3(M)={\mathbb Z}}$. Consider the generator of ${\pi_3(M)}$, namely ${\gamma: S^3\rightarrow M}$ which is of degree 1. This map induces isomorphisms on homology groups, by the definition of Hurewicz homomorphism, and thus by the above Corollary, ${M}$ is homotopy equivalent to ${S^3}$.${\Box}$

We introduce a new terminology: An ${n}$-homotopy sphere is an ${n}$-dimensional manifold that is homotopic to ${n}$-sphere. Using this terminology, we now know that any simply-connected closed 3-manifold is homotopy 3-sphere, and vice versa. This fact leads to the (seemingly) proper way to generalise the Poincaré conjecture:

Generalised Poincaré Conjecture(version1) Any homotopy ${n}$-sphere is homeomorphic to ${S^n}$.

We can think of another way of generalising the original conjecture. To get motivated, recall the definition of a manifold. The most important character of a manifold is that it is locally Euclidean. Now, a Euclidean space is a playground for analysis, and differentiation is the one of the most important (and most closely related to Topology) concept in analysis. So, yes. The direction of the generalisation is the structure of manifolds. Instead of continuity, we can think of differentiability. Also, motivated by polytope, one can think of piecewise-linear’ structure on a manifold, but we will not talk any details about it.

To describe this generalisation (including PL), we can adopt a terminology of a category. Category consists of data, called objects and morphisms. For example, we have a category consists of abelian groups with group homomorphisms, and a category consists of topological spaces with continuous maps(as morphisms). We are interested in manifolds’ here, but with various structures on it, we have the following three categories:

TOP: a category consists of topological manifolds with continuous maps.
DIFF: a category consists of smooth manifolds with smooth maps.
PL: a category consists of piecewise-linear manifolds with piecewise-linear maps.

Using these terminology, we can generalise the original Poincaré conjecture;

Generalised Poincaré Conjecture(version2) Any simply-connected closed 3-manifold is isomorphic to ${S^3}$ in each categories: TOP, DIFF, and PL.

Combining two generalisations, we get the Generalised Poincaré Conjecture;

The Generalised Poincaré Conjecture Any homotopy ${n}$-sphere is isomorphic to ${S^n}$ in each categories: TOP, DIFF, and PL.

On the next post (which I have no idea when will be posted), we can talk about whether the conjecture is true, and some of their proofs. Bye!

## Exotic 7-sphere

Apparently, I haven’t been posting anything for a while. It is mainly because I did not know what to write, as I do not have enough understanding on what I am currently trying to learn. With some desire on posting maths articles, I decided to write random things I have learnt in my previous studies.

During my undergraduate studies, I have been trying to understand the beautiful paper of J. Milnor, called On manifolds homeomorphic to 7-sphere’, so I am trying writing a little bit about it.

Everything starts with understanding manifolds. A manifold ${M}$ is a locally Euclidean topological space which is Hausdorff and second countable (here, we do not talk about a manifold with boundary). If every point of ${M}$ is locally homeomorphic to ${\mathbb R^n}$, then we say the dimension of ${M}$ is ${n}$. The invariance of domain tells us that if ${M}$ is connected, then the dimension has to be well-defined. When we simply say a manifold, we assume the connectedness.

There are many ways to understand’ an object, and one of the simplest, but maybe the most brutal and naive way is to classify (give the full list of ) objects. This being said, complete classification of all manifolds is maybe a dream for all topologists, which seems impossible to be done – the fundamental groups of manifolds have the word problem. Going down to earth, we can restrict manifolds to be, for example, compact or simply connected, and try the classification. For dimension 0,1 and 2, the full list of compact manifolds, at least orientable case, was known in 19th century, and so the next step was to classification for dimension 3.

This is where Poincaré conjecture comes in. In 1900, Henri Poincaré conjectures a statement equivalent to the following:

A 3-dimensional compact manifold (without boundary)
that has the same homology with ${S^3}$
must be homeomorphic to ${S^3}$.

After 4 years, in 1904, he found a counter-example of his previous conjecture by himself. It is called Poincaré homology sphere. (For those who are interested in this counter-example, you may find Eight faces of the Poincaré homology 3-sphere’ by R.C.~Kirby) Then he modified the previous one to get the famous conjecture, now known as Poincaré conjecture’. The conjecture is equivalent to the following:

A 3-dimensional compact manifold (without boundary)
that has the same fundamental group with ${S^3}$
must be homeomorphic to ${S^3}$.

Since the fundamental group of ${S^3}$ is trivial, i.e. ${S^3}$ is simply connected, the aforementioned conjecture is equivalent to,

Any simply connected 3-dimensional compact manifold (without boundary)
is homeomorphic to ${S^3}$.

This conjecture has been remained unsolved until the Russian mathematician, Grigori Perelman, proved the Thurston’s geometrisation conjecture which involves the Poincaré conjecture.

The Poincaré conjecture had led to a great improvement in the study of topology. Moreover, mathematicians generalised the conjecture (as generalisation is the closest friend of mathematician, together with coffee) to higher dimension and differentiable manifolds. The generalised Poincaré conjecture has also led to the great understanding of topology of that time. I will continue to talk about it in the next post.

## Recently

Although there were only a couple of posts for qualifying exams preparation, I took the test last weekend and then turns out to be ok- so there will be no more qualifying exams! yay!

Since then I have wasted all my life and hardly remember when I have been outside of my house – PLAYING GAMES ALL DAY LONG!

So, anyways, there might be no more qualifying exam related posts, and this blog is going to be used for archive interesting mathematical facts.

The plan for now is to learn more about stacks and differentiable stack, which I have learnt a little but haven’t studied much. Maybe I could start with posting things about those topics to help myself understand better.

It is getting cold over here and snow is always piled up. Maybe it is now time to be locked up inside the room and read maths. lol

## Qualifying Test-Linear Algebra

While preparing for the Qualifying test, I came up with an interesting (maybe only for me) question:

Question. Given a matrix ${A\in \mathfrak M_{n\times n}(\mathbb Z)}$, if ${Av=0}$ for a real vector ${v=[1,v_2,\cdots,v_n]\notin \mathbb Q^n}$, is there another(linearly independent) ${w\in {\mathbb R}^n}$ such that ${Aw=0}$

A naive and direct attempt would give an answer to this question:

Proof: Let ${a_i}$ be an ${i}$-th row vector of the matrix ${A}$. Then ${a_i\in \mathbb Q^n\subset {\mathbb R}^n}$. With the standard inner product in ${{\mathbb R^n}}$, we know that each ${a_i}$ is orthogonal to ${v}$. It directly follows that ${e_1=[1,0,\cdots,0]\notin \text{Span}\,(a_1,\cdots,a_n)}$ since ${\langle e_1,v\rangle=1\neq 0}$. Apply Gram-Schmidts process to ${e_1}$ over ${\{a_1,\cdots,a_n\}}$ to get a non-zero element ${w}$ (because ${e_1\notin \text{Span}\,(a_1,\cdots,a_n)}$) that is orthogonal to ${a_i}$ for all ${i}$. Notice that ${w\in \mathbb Q^n}$ since each ${a_i \in \mathbb Q^n}$ and ${\langle e_1,a_i\rangle \in \mathbb Q}$. Thus ${v}$ and ${w}$ are linearly independent and we get ${Aw=0}$. $\Box$

This is the first proof I could come up with. However, as I kept thinking about this question, I could get another proof which can be useful in many other ways. I start with the following observation(lemma):

Lemma 1 Suppose we have two fields ${F_1}$ and ${F_2}$ such that ${F_1\subset F_2}$ as a ring, and two ${n}$-dimensional vector spaces ${V_1}$ and ${V_2}$ over ${F_1}$ and ${F_2}$, respectively. That is, ${V_i\cong F_i^n}$ for ${i=1,2}$. Suppose, furthermore, that there is a basis ${\mathfrak B=\{e_1,\cdots,e_n\}}$ of ${V_1}$ which, under the inclusion ${F_1\subset F_2}$, is again a basis of ${V_2}$. Then any set of linearly independent vectors ${\{v_1,\cdots,v_k\}}$ in ${V_1}$ is still linearly independent in ${V_2}$.

Proof: Suppose there are ${a_1,\cdots,a_k \in F_2}$ such that ${\sum_{j=1}^{k} a_j v_j=0}$. Since ${v_j\in V_1}$, we may write ${v_j=\sum_{i=1}^n b_{ij}e_i}$. Then

$\displaystyle \sum_{j=1}^k a_j v_j=\sum_{i=1}^n\sum_{j=1}^k a_j b_{ij}e_i=\sum_{i=1}^n\left (\sum_{j=1}^k a_j b_{ij}\right )e_i=0$

and therefore by the fact that ${\mathfrak B}$ is again a basis of ${V_2}$, we have

$\displaystyle \sum_{j=1}^k a_j b_{ij}=0$

for all ${i}$. Let ${B=(b_{ij})}$ be an ${n\times k}$ matrix and ${a=(a_1,\cdots,a_k)\in F_2^k}$, then the above equation becomes

$\displaystyle Ba=0.$

From the fact that ${\{v_1,\cdots,v_k\}}$ is linearly independent in ${V_1}$, ${B}$ is of rank ${k}$ as a matrix in ${F_2}$ because the RREF form reduced in ${F_1}$ is already the RREF form in ${F_2}$, hence we get ${a=(a_1,\cdots,a_k)=0}$. In other words, ${\{v_1,\cdots,v_k\}}$ is linearly independent in ${V_2}$. $\Box$

From this lemma, we have the following corollary.

Corollary 2 Under the setting of Lemma 1, if there is a ${F_2}$-linear operator ${T:V_2\rightarrow V_2}$ where ${S=T|_{V_1}}$ is ${V_1}$ invariant as a ${F_1}$-linear operator, then ${\dim \ker T=\dim \ker S}$ and ${\dim \mathrm{im}\, T= \dim \mathrm{im}\, S}$.

Proof: For ${v\in V_1\subset V_2}$, ${Sv=0}$ implies ${Tv=0}$. Thus, as a set, ${\ker S\subset \ker T}$. By the above lemma, we get ${\dim \ker S \leq \dim \ker T}$. Similarly, we get ${\dim \mathrm{im}\, S \leq \dim \mathrm{im}\, T}$. Then by the rank theorem, applied independently for ${S}$ and ${T}$, we get ${\dim \ker T=\dim \ker S}$ and ${\dim \mathrm{im}\, T= \dim \mathrm{im}\, S}$. $\Box$

Using this corollary, we get another proof for the Question.

Proof: Set ${F_1=\mathbb Q}$, ${F_2=\mathbb R}$, ${V_1=\mathbb Q^n}$, ${V_2=\mathbb R^n}$ and ${e_i}$ to be the standard basis of ${{\mathbb Q}^n}$. Notice that ${T=L_A: v\mapsto Av}$ can be restricted to ${\mathbb Q^n}$ and get ${S}$ in the corollary. Since ${v\in \ker T}$ with ${v\notin {\mathbb Q}^n}$, there exist ${w\in ker S\subset {\mathbb Q}^n}$ by the dimension check. Since ${v}$ and ${w}$ are linearly independent, we are done. $\Box$

Posted in Maths, Qualifying exams | Tagged | 3 Comments

## Convergence of measurable functions

One can say that Mathematical Analysis is a theory of limits. For example, integration and differentiation cannot be described without limits. Also, approximation is a concept that cannot be apart from limits. Among all those limits, one of the most basic concept that comes with limits would be a sequence. When there is a sequence, either a sequence of numbers or a sequence of functions, one of the main issues is convergence. That is, we would like to observe where this infinitely many “items” lead to.

In this post, I would like to write about various types of convergences of a sequence of measurable functions. Their definitions and some relations are going to be presented here. All things written here is very basic but it might help you to organise each concepts clearly.

Throughout this post, ${X}$ is a measurable (metric) space with a positive measure ${\mu}$, and ${\{f_n\}_n}$ is a sequence of non-negative measurable functions.

Definition 1 A space ${X}$ is called measurable if ${X}$ is equipped with a collection ${\mathfrak M}$ of subsets of ${X}$ such that ${\emptyset, X \in \mathfrak M}$, ${E\in \mathfrak M}$ implies ${E^c \in \mathfrak M}$, and ${\{E_i\}_{i\in {\mathbb Z}}\subset \mathfrak M}$ implies ${\cup_{i\in {\mathbb Z}}E_i\in \mathfrak M}$. We call ${\mathfrak M}$ a ${\sigma}$-algebra and members of ${\mathfrak M}$ measurable sets.

A positive measure ${\mu}$ on ${X}$ is a function from ${\mathfrak M}$ to non-negative (extended) real number which satisfies countably additivity: If ${\{E_i\}\subset \mathfrak M}$ is a collection of mutually disjoint measurable sets, then

$\displaystyle \mu(\cup E_i)=\sum \mu(E_i).$

A function ${f:X\rightarrow {\mathbb C}}$ is called measurable if ${f^{-1}(V)\in \mathfrak M}$ for all open sets ${V\subset {\mathbb C}}$.

Fundamentally, we would like to talk about complex-valued measurable functions ${f_n}$, but it is sufficient to consider non-negative measurable functions. For any complex-valued function ${f(x)=u(x)+i\cdot v(x)}$ is measurable if and only if real valued functions ${u(x)}$ and ${v(x)}$ are measurable, and we know that a real valued function ${g(x)}$ splits into two non-negative functions ${g^+(x)}$ and ${g^-(x)}$ where ${g(x)=g^+(x) - g^-(x)}$.

Here is the most basic convergence of sequence of measurable functions.

Definition 2 (pointwise almost everywhere convergence) We say a sequence ${\{f_n\}}$ converges to a function ${f}$ in pointwise almost everywhere sense if there is a set ${E\subset X}$ with ${\mu(E)=0}$ and for each ${x\in X\backslash E}$, given any ${\epsilon>0}$, there is ${N\in {\mathbb Z}_{> 0}}$ so that

$\displaystyle n>N \Rightarrow |f_n(x)-f(x)|<\epsilon.$

In this case, we denote it by ${f_n \rightarrow f \,(a.e.)}$.

In the spirit of (Lebesgue) integration, “almost everywhere” sense (discarding measure zero set) is reasonable as the integration of a measure zero set is always zero.

While the above convergence is checking the convergence of a sequence of numbers ${\{f_n(x)\}_n}$, for each ${x\in X\backslash E}$, there is a more measure-theoretic way to define a convergence. When we have a sequence of functions ${\{f_n\}_n}$ which converges to ${f}$ in the sense of measure, it is very natural to come up with the following:

$\displaystyle \mu(\{x: |f_n(x)-f(x)|>\epsilon\}\rightarrow 0,\quad n\rightarrow \infty.$

for all ${\epsilon>0}$.

The above convergence “in the sense of measure” is equivalent to the following definition.

Definition 3 (In measure convergence) We say that a sequence ${\{f_n\}_n}$ of measurable functions converges in measure if for any ${\epsilon>0}$, there is ${N\in {\mathbb Z}_{>0}}$ s.t. ${n>N}$ implies

$\displaystyle \mu(\{x: |f_n(x)-f(x)|>\epsilon\})<\epsilon.$

There is one more convergence I would like to mention: ${L^p}$-convergence. Unlike other convergence, we require ${f_n}$ to be more than measurable. In order to describe the ${L^p}$-convergence, let me briefly describe what ${L^p}$ space is. We assume that ${p\geq 1}$.

Suppose we have a measurable function ${f:X\rightarrow {\mathbb C}}$ (or ${{\mathbb R}}$). Since two functions ${x\mapsto |x|}$ and ${x\mapsto x^p}$ are continuous, their composition ${|f|^p}$ with measurable function ${f}$ is measurable. Finally, let ${\|f\|_p=\left [ \int_X |f|^p d\mu\right ]^{1/p}}$, and we obtain a new vector space, called ${L^p}$ space, consists of all measurable complex(or real) valued function ${f}$ with ${\|f\|_p<\infty}$. In fact, the functional ${\| \cdot \|_p}$ is a norm, called ${L^p}$-norm, and we may define a convergence with respect to this norm.

Definition 4 (${L^p}$-convergence) We say that a sequence ${\{f_n\}_{n\in {\mathbb Z}_{>0}}}$ of functions, in which ${f_n \in L^p}$ for all ${n\in {\mathbb Z}_{>0}}$, converges to ${f}$ in ${L^p}$-norm if for each ${\epsilon>0}$, there is ${N\in {\mathbb Z}_{>0}}$ such that ${n>N}$ implies

$\displaystyle \|f_n-f\|_p<\epsilon.$

Now we have all the ingredients for the post. We have (1)Almost Everywhere Pointwise Convergence(AEPC), (2)In Measure Convergence(MC) and (3)${L^p}$-Convergence(LPC) for ${p\geq 1}$. The objective of this post is to state the relation between those three convergences.

Theorem 5 For a measurable (metric) space ${X}$ with a positive measure ${\mu}$, we have the following relations:

1. AEPC implies MC if ${\mu(X)<\infty}$.
2. AEPC does NOT imply LPC. Even more, the existence of LPC subsequence (of an AEPC sequence) is not guaranteed.
3. MC sequence has a subsequence that is AEPC.
4. MC does NOT imply LPC.
5. LPC sequence has a subsequence that is AEPC.
6. LPC implies MC.

One of Littlewood’s principles says that AEPC is nearly almost everywhere uniformly convergent. In more precise term, we have the following theorem.

Theorem 6 (Egoroff’s Theorem) Let ${\{f_n\}_{n\in {\mathbb Z}_{>0}}}$ be a sequence of measurable functions on a measurable space ${X}$ with ${\mu(X)<\infty}$ which pointwisely converges almost everywhere to a measurable function ${f}$. Then for any ${\epsilon>0}$ the sequence ${\{f_n\}}$ uniformly converges almost everywhere to ${f}$ on some measurable set ${E\subset X}$ where ${\mu(X\backslash E)<\epsilon}$. Moreover, this measurable set can be taken to be a closed set.

The consequence of Egoroff’s theorem is the following: Notice that the pointwise limit of a sequence of measurable functions is again measurable (if it exists almost everywhere). If we restrict the sequence of functions to be continuous, we still get measurable function instead of continuous function. However, thanks to Egoroff’s theorem, we can expect the pointwise limit of a sequence of continuous functions to be continuous except on an arbitrary small measure. This is a huge advantage as continuous functions are more familiar for most people and behave more decently.

I should probably sketch the proofs of two theroems above, but I will postpone it for a while. I would edit this post later to write down the sketch of proofs.