Math Challenge - April 2021

[mathwonk]

E.g. one version of the nullstellensatz is that every maximal ideal in C[X1,...,Xn] is the ideal of a single point, say p = (a1,...,an), hence has form (X1-a1,...Xn-an). You might think this is obvious, since a maximal ideal must correspond to a minimal variety, and surely one point is a minimal variety. But no, the empty set is even smaller! so the whole point of the nullstellensatz is to prove that the variety of every proper ideal is non empty.

 

[Office_Shredder]

Yeah I think the intersection is straightforward:

If $X=V(I)$ and $Y=V(J)$ are varieties, then consider $V(I+J)$. If $f+g\in I+J$, with $f\in I$ and $g\in J$, then for all $x\in X\cap Y$, we have $f(x)+g(x)=0$. So $X\cap Y \subset V(I+J)$. To get the other direction, If, say, $x\notin X\cap Y$, suppose without loss of generality that $x\notin X$. Then there exists $f\in I$ such that $f(x)\neq 0$. Hence $f+0\in I+J$ is a polynomial that is not zero on $x$, so $V(I+J)\subset X\cap Y$.

When I get a chance to do it on my laptop I can try to type up a more complete solution to #4.

 

[fresh_42]

When I get a chance to do it on my laptop I can try to type up a more complete solution to #4.

Here is mine:

Spoiler: #4

Assume $Y$ is irreducible and $f,g\in \mathcal{O}_\mathbb{C}(\mathbb{A}^n)$ such that $f\cdot g\in I(Y).$ Then $V(fg)=V(f)\cup V(g)$ and
$$
Y=(V(f)\cap Y)\cup (V(g)\cap Y).
$$
Since $Y$ is irreducible, we have w.l.o.g. $Y=V(f)\cap Y,$ i.e. $Y\subseteq V(f)$ and $f\in I(Y).$

Now let $J:=I(Y)$ be a prime ideal. Let $V(J)=Y_1\cup Y_2.$ Thus $J=I(Y_1\cup Y_2)=I(Y_1)\cap I(Y_2).$ Assume $J\neq I(Y_1),I(Y_2).$ Then there are $f_i\in I(Y_i)-J.$ Since
$$
f_1f_2\in I(Y_1)\cdot\mathcal{O}_\mathbb{C}(\mathbb{A}^n)\cap \mathcal{O}_\mathbb{C}(\mathbb{A}^n)\cdot I(Y_2)=I(Y_1)\cap I(Y_2)=J
$$
is a prime ideal, we have that either $f_1\in J$ or $f_2\in J$ contradicting the choice of $f_1,f_2.$ Hence w.o.l.g. we may assume $I(Y)=J=I(Y_1).$ This implies $Y=V(I(Y))=V(I(Y_1))=Y_1$ and $V(J)$ is irreducible.

 

[mathwonk]

#4 Spoiler:

One can approach it as follows:
1. If X and Y are varieties, then X is contained in Y if and only if I(Y) is contained in I(X).
2. If S and T are collections of polynomials, then V(S)intersect V(T) = V(S union T).

Cor: A variety X is irreducible if and only if I(X) is prime.
pf: By 1., If a variety X = Y unionZ, where Y and Z are varieties, neither contained in the other, then there exist polynomials f, g with f vanishing on X but not Y, and g vanishing on Y and not X, hence fg belongs to I(X) even though neither f nor g does. So I(X) is not prime. Conversely, if I(X) is not prime and f.g belongs to I(X) but neither f nor g do, then X = (XintersectV(f)) union (X intersect V(g)), shows X is reducible, by 2.

this is essentially the union of office shredder and fresh 42's solutions.

 

[Infrared]

Since no one has touched the third problem, I'll give a skeleton of a proof that someone might like to fill in.

It should be clear that (a) and (b) are equivalent. To show that (b) is equivalent to (c), you want to show that there is an antipodal map $S^n\to S^{n-1}$ if and only if there is a map $B^n\to S^{n-1}$ that is antipodal on the boundary.

Given an antipodal map $S^n\to S^{n-1}$, you can restrict to the northern hemisphere to get a map $B^n\to S^{n-1}$ which is antipodal on the boundary (equator of the original $S^n$). Conversely if there is a map $B^n\to S^{n-1}$ that is antipodal on the boundary, you can glue two copies of $B^n$ along their boundary (with a twist) to get a sphere $S^n$ and the maps glue to give an antipodal map $S^n\to S^{n-1}.$

To show Brouwer's fixed point theorem when these all hold: If $f:B^n\to B^n$ has no fixed point, then consider the map $B^n\to \partial B^n=S^{n-1}$ defined by taking $x$ to the intersection point of the ray from $f(x)$ to $x$ and the boundary $S^{n-1}.$ This is a continuous map $B^n\to S^{n-1}$ which is the identity on the boundary. Then compose with the antipodal map to contradict part (c).

Edit:

I think there's a typo in the problem statement.

3. A function $f\, : \,S^k\longrightarrow X$ is called antipodal if it is continuous and $f(-x)=-f(x)$ for all $x\in S^k$ and any topological space $X.$

What does $-f(x)$ mean when $f(x)$ is an element of an arbitrary topological space $X$. I guess you want $X$ to be a (topological) vector space or abelian group or something.

 

[fresh_42]

Edit:

I think there's a typo in the problem statement.
What does $-f(x)$ mean when $f(x)$ is an element of an arbitrary topological space $X$. I guess you want $X$ to be a (topological) vector space or abelian group or something.

Thanks for mentioning it. Yes, that should have been $\subseteq \mathbb{R}^m$. I made a mistake when putting all those different functions of the theorem in one definition. Corrected.

 

[Office_Shredder]

Edit: I tried to generalize the notion of antipodal but what I wrote is obviously wrong. Will rethink it.

 

[Not anonymous]

I am unsure about the correctness of the answer for the general case as I find it difficult to validate it for $n > 3$ and some of the properties of determinants I use in the proof are ones which I have only seen in supplementary books/sections that were hardly ever covered in main curriculum and so this might be the first time I am trying to use those.

Spoiler: Question 14

For $n=1$, i.e. the case of single-digit numbers, it is fairly obvious that $det(N) = N$ as the determinant of a 1x1 matrix equals the value of the one element. Hence $f(1) = \sum_{N=0}^9 det(N) = \sum_{N=0}^9 N = \frac {9 \times 10} {2} = 45$.

For $n=2$, let us assume that the decimal representation of $N$ is $abcd$, where $a \in \{1, 2, .., 9\}$ and $b, c, d \in \{0, 1, 2, .., 9\}$. In this case, $det(N) = a \times d - b \times c$ and $f(n) = \sum_{N=1000}^{9999} det(N) = \sum_{a=1}^{9} \sum_{b=0}^{9} \sum_{c=0}^{9} \sum_{d=0}^{9} (a \times d - b \times c) =$
$\sum_{a=1}^{9} \sum_{d=0}^{9} (a \times d \times \sum_{b=0}^{9} \sum_{c=0}^{9} 1) - \sum_{b=0}^{9} \sum_{c=0}^{9} (b \times c \times \sum_{a=1}^{9} \sum_{d=0}^{9} 1) =$
$\sum_{a=1}^{9} \sum_{d=0}^{9} (a \times d \times 100) - \sum_{b=0}^{9} \sum_{c=0}^{9} (b \times c \times 90) = 100 \times 45^2 - 90 \times 45^2 = 10 \times 45^2 = 20250$.

For $n \geq 3$, let us assume that the decimal representation of $N$ is $a_{1,1}a_{1,2}...a_{1,n}a_{2,1}...a_{n,1}...a_{n,n}$, where $a_{1,1} \in \{1, 2, .., 9\}$ and $a_{1,2}, ..., a_{n,n} \in \{0, 1, 2, .., 9\}$. The corresponding matrix representation is
$$\begin{bmatrix}
a_{1, 1} & a_{1, 2} & .. & a_{1,n} \\
a_{2, 1} & a_{2, 2} & .. & a_{2,n} \\
... \\
a_{n, 1} & a_{n, 2} & .. & a_{n,n}
\end{bmatrix}$$ and $$f(n) = \sum_{N=10^{n^2 - 1}}^{10^{n^2} - 1} det(N)$$.

For every distinct value of $N$ in the summation, one of the following conditions must be true:
(a) $det(N) = 0$
(b) $det(N) \neq 0$. This implies that that last $n$-digit sequence in the decimal representation of $N$ is not the same as the sequence formed by the preceding $n$ digits, i.e. $\begin{bmatrix}a_{n-1,1} & a_{n-1, 2} & .. & a_{n-1, n} \end{bmatrix} \neq \begin{bmatrix}a_{n,1} & a_{n, 2} & .. & a_{n, n} \end{bmatrix}$, since otherwise $det(N)$ would be 0 (since determinant of a matrix with 2 rows being the same is zero)

Values of $N$ that meet the condition (a) do not affect the sum and hence can be dropped from the summation. The remaining values in the summation would meet condition (b) and can be grouped uniquely to form pairs $(N_{1}, N_{2})$ such that the matrix representation of $N_{2}$ is derived from the matrix representation of $N_{1}$ by interchanging the last 2 rows, and there will always be exactly one unique $N_{2}$ for any given $N_{1}$. Since the determinant of a matrix $A$ derived by interchanging 2 adjacent rows of a matrix $B$ is the negative of the determinant of $B$, the above pairing of values in the summation implies that every positive-valued determinant $det(N_1)$ in the summation will be canceled by its negative counterpart $det(N_2)$. This means that the summation value equals zero.

Hence $f(n) = \begin{cases} 10^{(n-1)} \times 45^{n}, & n \leq 2 \\ 0, & n \gt 2 \end{cases}$

 

[fresh_42]

I am unsure about the correctness of the answer for the general case as I find it difficult to validate it for $n > 3$ and some of the properties of determinants I use in the proof are ones which I have only seen in supplementary books/sections that were hardly ever covered in main curriculum and so this might be the first time I am trying to use those.

Spoiler: Question 14

For $n=1$, i.e. the case of single-digit numbers, it is fairly obvious that $det(N) = N$ as the determinant of a 1x1 matrix equals the value of the one element. Hence $f(1) = \sum_{N=0}^9 det(N) = \sum_{N=0}^9 N = \frac {9 \times 10} {2} = 45$.

For $n=2$, let us assume that the decimal representation of $N$ is $abcd$, where $a \in \{1, 2, .., 9\}$ and $b, c, d \in \{0, 1, 2, .., 9\}$. In this case, $det(N) = a \times d - b \times c$ and $f(n) = \sum_{N=1000}^{9999} det(N) = \sum_{a=1}^{9} \sum_{b=0}^{9} \sum_{c=0}^{9} \sum_{d=0}^{9} (a \times d - b \times c) =$
$\sum_{a=1}^{9} \sum_{d=0}^{9} (a \times d \times \sum_{b=0}^{9} \sum_{c=0}^{9} 1) - \sum_{b=0}^{9} \sum_{c=0}^{9} (b \times c \times \sum_{a=1}^{9} \sum_{d=0}^{9} 1) =$
$\sum_{a=1}^{9} \sum_{d=0}^{9} (a \times d \times 100) - \sum_{b=0}^{9} \sum_{c=0}^{9} (b \times c \times 90) = 100 \times 45^2 - 90 \times 45^2 = 10 \times 45^2 = 20250$.

For $n \geq 3$, let us assume that the decimal representation of $N$ is $a_{1,1}a_{1,2}...a_{1,n}a_{2,1}...a_{n,1}...a_{n,n}$, where $a_{1,1} \in \{1, 2, .., 9\}$ and $a_{1,2}, ..., a_{n,n} \in \{0, 1, 2, .., 9\}$. The corresponding matrix representation is
$$\begin{bmatrix}
a_{1, 1} & a_{1, 2} & .. & a_{1,n} \\
a_{2, 1} & a_{2, 2} & .. & a_{2,n} \\
... \\
a_{n, 1} & a_{n, 2} & .. & a_{n,n}
\end{bmatrix}$$ and $$f(n) = \sum_{N=10^{n^2 - 1}}^{10^{n^2} - 1} det(N)$$.

For every distinct value of $N$ in the summation, one of the following conditions must be true:
(a) $det(N) = 0$
(b) $det(N) \neq 0$. This implies that that last $n$-digit sequence in the decimal representation of $N$ is not the same as the sequence formed by the preceding $n$ digits, i.e. $\begin{bmatrix}a_{n-1,1} & a_{n-1, 2} & .. & a_{n-1, n} \end{bmatrix} \neq \begin{bmatrix}a_{n,1} & a_{n, 2} & .. & a_{n, n} \end{bmatrix}$, since otherwise $det(N)$ would be 0 (since determinant of a matrix with 2 rows being the same is zero)

Values of $N$ that meet the condition (a) do not affect the sum and hence can be dropped from the summation. The remaining values in the summation would meet condition (b) and can be grouped uniquely to form pairs $(N_{1}, N_{2})$ such that the matrix representation of $N_{2}$ is derived from the matrix representation of $N_{1}$ by interchanging the last 2 rows, and there will always be exactly one unique $N_{2}$ for any given $N_{1}$. Since the determinant of a matrix $A$ derived by interchanging 2 adjacent rows of a matrix $B$ is the negative of the determinant of $B$, the above pairing of values in the summation implies that every positive-valued determinant $det(N_1)$ in the summation will be canceled by its negative counterpart $det(N_2)$. This means that the summation value equals zero.

Hence $f(n) = \begin{cases} 10^{(n-1)} \times 45^{n}, & n \leq 2 \\ 0, & n \gt 2 \end{cases}$

The zero in case $n>2$ can shortly be explained by:
If $n>2$ then all determinants appear as positive and equal negative value in the sum: Fix all but the last two columns $(c_{n-1},c_{n}).$ Then $(c_{n},c_{n-1})$ is in the sum, too, but of opposite sign.

 

[Office_Shredder]

That's an awesome solution for #14!

 

[mathwonk]

For the interested, the next two posts will strengthen Problem #4 to show the correspondence there sets up a bijection between all prime ideals of C[X1,...,Xn] and all irreducible affine subvarieties of C^n.

Let k be an uncountable field, e.g. the real or complex numbers, and consider the polynomial ring k[X] and its quotient field k(X).

1. Show that as, a vector space over k, k(X) has uncountable dimension, by showing the set of fractions 1/(X-a) for all a in k, is an uncountable linearly independent set.

2. Now consider the ring k[X,Y] and show it has countable vector dimension over k.

3. Now let M be a maximal ideal of k[X,Y], and let m be the intersection of M with k[X]. Then we have an inclusion of the domain k[X]/m into the field k[X,Y]/M. Prove the prime ideal m is not (0).

4. Deduce that m, hence also M, contains a non-trivial irreducible polynomial P(X) with coefficients in k.

5. Using a similar argument for k[Y], deduce that k[X,Y]/M is an algebraic field extension of k. [This is actually true for any field k, and is called Zariski's nullstellensatz. It is proved for infinite fields in section 2.2 of the book Introduction to Algebraic Geometry, by Justin Smith, and for all fields in the first section of Mumford's red book, albeit with a slightly sketchy proof of the normalization lemma. Miles Reid gives more details in his Undergraduate Commutative Algebra. So perhaps curiously, the larger the field k is, the easier the proof.]

6. In particular if k is algebraically closed and uncountable, the inclusion of k into k[X,Y]/M is an isomorphism, and M has form M = (X-a,Y-b) for some elements a,b in k. In particular, if C is the complex field, the correspondence between points of C^2 and maximal ideals of C[X,Y] is bijective.

7. Conclude that if I is any ideal in C[X,Y], then V(I) is non empty.

A straightforward induction argument proves the same results for C[X1,...,Xn]. In particular the procedure in problem #4, sets up a one-one correspondence between points of C^n and maximal ideals of C[X1,...,Xn].

 

[mathwonk]

Nullstellensatzes:

We know (“weak” nullstellensatz) that if C is the complex numbers then for every proper ideal J of C[X] = C[X1,…,Xn], V(J) is a non empty affine subvariety of C^n. If W is any affine subvariety of C^n, with ideal I(W) and affine ring C[X]/I(W), then it follows that for any proper ideal J of this quotient ring, V(J) is a non empty affine subvariety of W. I.e. the weak nullstellensatz implies there is a bijective correspondence between points of an affine variety and maximal ideals of its affine ring. Thus the points of an affine variety are determined once we know the associated affine ring.

So we can define an abstract affine variety as a set V together with a finitely generated C-algebra A of C-valued functions on V, such that associating a point p of V with the ideal I(p) of functions in A vanishing at p, sets up a 1-1 correspondence between points of V and maximal ideals in A. Then choosing a set of n algebra generators for A defines an embedding of V as an affine subvariety of C^n. Conversely, any affine subvariety V of C^n has an affine algebra of functions A = C[X1,...,Xn]/I(V), whose maximal ideals correspond to the points of V, and which is generated by the restrictions of the variables X1,..,Xn to V. Thus an embedded affine variety is just an abstract affine variety plus a choice of a finite set of algebra generators for its affine ring.

Now let f be any non constant polynomial in C[X], and V(f) the affine hypersurface it defines, and U(f) = C^n-V(f) the complementary open subset of C^n. We claim that this open set U(f) is isomorphic to a closed affine subvariety W of C^(n+1). Namely define W = V(1-f(X).Y), by the polynomial 1-f(X).Y, in the ring C[X,Y] = C[X1,…,Xn,Y] of polynomials in n+1 variables. Then the projection C^(n+1)—>C^n sending (a1,…,an,b) to (a1,…,an), maps the affine variety W bijectively onto U(f), with inverse map sending a = (a1,…,an) to (a,1/f(a)), which is defined since f does not vanish on U(f). Thus the affine ring of W, namely C[X,Y]/(1-f(X).Y) is isomorphic to the ring C[X][1/f], obtained from C[X,Y] by setting Y equal to 1/f. I.e. as an abstract affine variety, U(f) has affine ring C[X][1/f]. In particular the weak nullstellensatz holds for U(f) and this ring, so that every proper ideal of this ring has a common zero lying in U(f).

Now let J be any ideal of C[X] and assume that f vanishes on V(J), i.e. that V(f) contains V(J), or equivalently, that V(J) does not intersect U(f). This means that the ideal J of C[X], generates an ideal of C[X][1/f] that has no zeroes in U(f), so by the weak nullstellensatz, it is not a proper ideal. Thus there is a finite linear combination of elements of J, with coefficients in C[X][1/f] that equals 1. Multiplying out the denominators by some high power of f, we get a linear combination of elements of J, with coefficients in C[X] equal to some positive power f^r, of f.

We conclude that if J is any ideal of C[X], and f is any polynomial of C[X] that vanishes on V(J), then some positive power of f belongs to J, i.e. f is in the “radical” of J. Thus I(V(J)) = rad(J). (We have proved the inclusion from left to right; the other inclusion is easy.) This is called the “strong” nullstellensatz.

Using this, conclude that in problem #4, there is actually a one-one correspondence between all irreducible affine subvarieties of C^n and all prime ideals of C[X1,...,Xn].

[Justin Smith uses the same algebraic manipulation ("the trick of Rabinowitsch") in his book, to deduce the strong from the weak nullstellensatz, but I have tried to motivate this well known trick by explaining that U(f) is isomorphic to an affine variety with ring k[X][1/f]. His argument is thus shorter, but to me less motivated.]

 

[mathwonk]

SPOILER: Hint for problem #9:

Try to show every continuous map from the circle to the sphere, say taking the point 1 to the north pole, is homotopic to a constant. As a lemma, prove such a map is homotopic to a constant if it is not surjective. Then, prove you can subdivide the circle into a finite number of arcs, such that your map is homotopic to a map that takes each arc into an arc on the sphere.

 

[fresh_42]

SPOILER: Hint for problem #9:

Try to show every continuous map from the circle to the sphere, say taking the point 1 to the north pole, is homotopic to a constant. As a lemma, prove such a map is homotopic to a constant if it is not surjective. Then, prove you can subdivide the circle into a finite number of arcs, such that your map is homotopic to a map that takes each arc into an arc on the sphere.

One can also apply a famous theorem from algebraic topology. In that case, the proof consists of creating a suitable setup in order to apply the theorem.

 

[mathwonk]

yes, I was trying to avoid invoking the machinery of simplicial approximation. One could sort of combine those ideas by showing the approximating spherical arcs can be made to consist entirely of pieces of arcs of latitude or longitude. One could then further subdivide the sphere into simplices containing those arcs, obtaining a simplicial approximation. My goal was to make it accessible to someone who has not studied too much algebraic topology. Of course one needs to be familiar with homotopy even to grasp the statement. But I think one only needs to understand the (words in the) definition of the fundamental group to do this problem.

 

[Infrared]

@mathwonk I think @fresh_42 was referencing Van-Kampen. I also interpreted the question to be done directly (in which case the job is, as you noted, to find a nice way to homotope a loop to make it non-surjective).

 

[mathwonk]

Ah yes. I am not too familiar with Van Kampen's theorem myself. But actually now I realize the proof of the surjectivity part of Van Kampen seems not much more complicated, intuitively at least, than the direct argument I had in mind, and that suffices here. I guess I tend to like thinking through an argument from the ground up more than wielding a big hammer. But thank you for referring me to this one, since this has enabled me to finally see the simplicity of, and hence appreciate, at least the easy part of this hammer a little better. I.e. the basic idea is just that given any map of the circle into the sphere, we can subdivide the circle into arcs, so that each arc maps entirely into either the (fattened) upper hemisphere or the (fattened) lower one. Then for Van Kampen you just have to finagle the base points, to make each image of an arc (homotopic to) a union of images of loops with the given base point (this time chosen to lie on the equator)...Ah yes, and you need path connectedness of the equator, so that you can choose those loops so that their composition is still homotopic to the original map. Thank you again! Now what was a mysterious theorem for me is less so.

 

[fresh_42]

Then for Van Kampen you just have to finagle the base points, to make each image of an arc (homotopic to) a union of images of loops with the given base point (this time chosen to lie on the equator)...Ah yes, and you need path connectedness of the equator, so that you can choose those loops so that their composition is still homotopic to the original map.

I'm not sure whether this works for your basic argument as well, but Seifert - van Kampen has the advantage, that the same argument covers the slightly more general case:

If $X=U\cup V$ with open sets $U,V$ and $U\cap V$ is path connected, then
$$
\pi_1(U;x)=\pi_1(V;x)=\{e\}\Longrightarrow \pi_1(X;x)=\{e\}.
$$

 

[Office_Shredder]

Here's my attempt at #9, which does not use any machinery from algebraic topology, or even stereographic projection.

First, some algebraic topology background to explain what the question is asking for. Feel free to skip this if you are familiar with the context of the question.

Spoiler

$\pi(S^n,x)$ is the fundamental group or "loop group" of loops of $S^n$ based at a point $x$. A loop based at a point $x$ is just a curve which starts and ends there, i.e. a continuous function $f:[0,1]\to S^n$ such that $f(0)=f(1)=x$. We can define an equivalence relationship on loops via homotopy, which is a process of continuously deforming one curve into another. Suppose $f$ and $g$ are curves on a space $X$ with $f(0)=g(0)$ and $f(1)=g(1)$, then we say these are homotopic if there exists $F(t,s):[0,1]\times [0,1]\to X$ a continuous function such that $F(0,s)=f(0)=g(0)$ for all $s$, $F(1,s)=f(1)=g(1)$ for all $s$, and $F(t,0)=f(t)$ for all $t$, and $F(t,1)=g(t)$ for all $t$ (I apologize at this point, I can never remember the convention of whether the first or second variable is held fixed for each of these, so this might not be the standard convention).

If this definition seems a bit uninuitive, consider any two curves $f$ and $g$ on $\mathbb{R}^3$ that start and end at the same point. Then consider $F(t,s) = f(t)s + g(t)(1-s)$. This matches the technical definition of a homotopy, and intuitively it's just slowly transforming the curve $f$ into the curve $g$ by interpolating between them (try drawing some curves, and then draw $F(t,1/2)$ and it will be obvious probably).

It turns out that if you take the set of all loops that are based at a specified point, and you consider the set of equivalence classes, these form a group. Multiplication is just concatenating loops, the multiplicative inverse is going around a loop in the opposite direction, and the identity is the constant loop that just always sits at a point.

This is is not actually necessary for #9, #9 just says that the space of equivalence classes of loops just has one equivalence class, which is the constant loop that always sits at a point. So all we need to do is prove that there exists a homotopy to a constant loop, and the group structure is irrelevant to the actual question.

In $\mathbb{R}^n$ this is obvious, given a loop $f$ at a point $x$, we can just define $F(t,s) = f(t)s + (1-s)x$, and $F(1,s)=x$ for all $s$ demonstrates that $f$ is homotopic to the constant loop for any loop.

On $S^1$ this turns out to not be true, a loop $f(t) = (\cos(2\pi t),\sin(2\pi t))$ is *not* homotopic to a constant loop. This fact I think is not exactly trivial to prove, but is not a super deep theorem.

For $S^n$ for $n\geq 2$, we wish we could just do the same thing as in $\mathbb{R}^n$: let's WLOG define the base of our loop to be the north pole, then we want to just interpolate our loop back to the north pole of the sphere. The problem is that for the south pole of the sphere, it's not obvious which direction we are supposed to interpolate. On $S^1$ this is impossible to get around but in higher dimensions we can just kind of pick an arbitrary direction to move it as long as we are careful enough (for simple curves on $S^2$ you should be able to imagine which way you are supposed to slide the south pole yourself).

Now we construct a homotopy sending a loop to the constant loop assuming it does not pass through the opposite point of where it is based.

Spoiler

Suppose the loop is based at $(0,0,0,...,0,1)$ and does not pass through $(0,0,...,0,-1)$. Then we can write any point that's not either of these poles uniquely as $\sin(\theta)x + \cos(\theta) y$ for $x=(x_1,...,x_{n-1},0)\in S^n$ and $y=(0,0,..,0,1)$ where $\theta \in (-\pi,0)$. If $\theta=-\pi$ then we get $-y$, and if $\theta=0$ then we get $y$ Then any curve $f(t)$ can be written as $\sin(\theta(t))x(t)+\cos(\theta(t))y$ where $x(t)$ is continuous in $t$ and $\theta(t)$ is continuous in $t$. The choice of $x(t)$ is unique as long as $f(t)\neq y$.

We can then define the homotopy
$$F(t,s) = \sin(\theta(t)(1-s))x(t)+\cos(\theta(t)s)y$$
for any point not equal to $\pm y$, and $F(t,s)=y$ whenever $f(t)=y$.

This is clearly continuous as it's a composition of continuous functions outside of $\pm y$. $f(t)\neq -y$ by assumption, so we don't even need to define what happens at that point. And this is continuous at $y$ as well by construction, if $f(t)=y$ then $F(t,s)=y$ for all $y$, and if $f(t)\neq y$, then $\lim_{s\to 1}F(t,s)=y$ for all $t$.

We are going to need a lemma, that $S^k$ is path connected for all $k\geq 1$

Spoiler

If $x\neq y$ are two points in $S^k$, then $\cos(t\pi/2) x + \sin(t\pi/2) y \in S^{n-1}$ for all $t \in [0,1]$ defines a path between the two points.

Now we show that any loop is homotopic to a loop not passing through the point opposite the base

Spoiler

Suppose $f$ is a loop with $f(0)=f(1)=y=(0,0,...,0,1)$. We want to show that $f$ is homotopic to a loop which does not pass through $-y$. Just to give some intuition, in the worst case scenario $f$ might be a space filling curve that passes through every point (multiple times? I have no idea. Uncountably many times? Maybe possible, I don't know), so we might be dealing with a curve arbitarily gross.

Let $D$ be a small disk around $-y$, of all points of the form $\sin(\theta)x-\cos(\theta) y$ where $x=(x_1,...,x_{n-1},0)\in S^n$, and $0<\theta < 0.1$. If $f(\tau)\in D$ for some fixed $\tau$, then there exists a largest $t_0<\tau$ such that $t_0\in \partial D$ is on the boundary of $D$, and a smallest $t_`1>\tau$ such that $t_1\in \partial D$. This is true since $\{t < \tau : f(t) \notin D \}$ has an upper bound $t_0$, and by continuity if $f(t_0)$ is in the interior of $D$, then we can find $\epsilon$ such that $f(t_0-\delta) \in D$ for all $\delta < \epsilon$, and hence $t_0-\epsilon$ is an upper bound of the set, and if $t_0$ lies outside of the closure of $D$, then similarly there exists $\epsilon$ such that $f(t+\delta)$ is outside of the closure of $D$ for all $0<\delta < \epsilon$, so $t_0$ is not an upper bound of the set.

Suppose that we had homotopies $F_{\alpha}(t,s)$ that are defined on $[t_{\alpha}^0,t_{\alpha}^1]\times [0,1]$ where $F_{\alpha}(t_{\alpha}^0,s)=f(t_{\alpha}^0)$ for all $s$, $F_{\alpha}(t_{\alpha}^1,s)=f(t_{\alpha}^1)$ for all $s$, and $F_{\alpha}(t,1)\in \partial D$ for all $t$#. Then we can define a homotopy $F$ on $[0,1]\times [0,1]$ by $F(t,s)=f(t)$ if $f(t) \notin D$, and $F(t,s)=F_{\alpha}(t,s)$ if $t\in [t_{\alpha}^0,t_{\alpha}^1]$. Then $F(t,1)$ is homotopic to $f$, and $F(t,1)\notin D$ for all $t$ as required.

We now construct $F_{\alpha}(t,s)$. We can write $f(t_{\alpha}^0) = \sin(.1)(x_{\alpha}^0,0) + \cos(.1)(0,0,...0,1)$ for $x_{\alpha}^0\in S^{n-1}$, and $f(t_{\alpha}^1)=\sin(.1)(x_{\alpha}^1,0) + \cos(.1)(0,0,...0,1)$ for $x_{\alpha}^1\in S^{n-1}$. By the preceding lemma about $S^{n-1}$ being path connected, there is some curve $h$ in $S^{n-1}$ such that $h(t_{\alpha}^0)=x_{\alpha}^0$ and $h(t_{\alpha}^1)=x_{\alpha}^1$. Let $g$ be the curve $\sin(0.1)(g(t),0)+\cos(0.1)(0,0,...,0,1)$ whch connects $f(t_{\alpha}^0)$ and $f(t_{\alpha}^1)$ with a curve entirely on $\partial D$. Then $F_{\alpha}(t,s) = f(t)\cos(s\pi/2)+g(t)\sin(s\pi/2)$ gives us the homotopy we require.

 

[Not anonymous]

Spoiler: Question 12

First we prove by contradiction that for any primitive Pythagorean triplet $(x, y, z)$ ($z^2$ being the sum on RHS), exactly one of either $x, y$ must be odd, while the other must be even.

• If both $x, y$ are even, then the RHS $z^2$ would also be even, in turn implying that $z$ is also even. This in turn implies that 2 is a common divisor of $x, y, z$, contradicting the requirement of primitiveness.
• If both of $x, y$ are odd, they can be written as $x=(4a + r_1), y=(4b + r_2)$ for some positive integers $a, b$ such that $r_1 = x \mod 4, r_2 = y \mod 4$ and $r_1, r_2 \in \{1, 3\}$. Then the Pythagorean sum becomes $z^2 = 4a^2 + 8ar_1 + r_{1}^{2} + 4b^2 + 8br_2 + r_{2}^{2} \\ \Rightarrow z^2 \mod{4} \equiv (r_1^2 + r_2^2) \mod{4} \equiv 2 \mod{4}$. This implies that $z^2$ is even but not a multiple of 4. But this would mean that there can be no integer solution for $z$ since an odd $z$ cannot given an even $^2$ and an even $z$ will mean $z^2 \equiv 0 \mod{4}$, again a contradiction.

Hence the only possibility is that one of $x,y$ is odd and the other is even. This implies $z^ = x^2 + y^2$ is odd and hence $z$ too must be odd.

Proof for the if part:
Given $u, v$ are coprime with $u > v$ and exactly one of them is odd. Let $x = u^2 - v^2, y = 2uv, z = u^2 + v^2$. $x^2 + y^2 = (u^4 + v^4 - 2u^2v^2) + 4u^2v^2 = (u^4 + v^4 + 2u^2v^2) = (u^2 + v^2)^2$
$\Rightarrow x^2 + y^2 = z^2$. Thus $x, y, z$ form a Pythagorean triplet. To prove that it is a primitive triplet, let us assume that on the contrary $x, y$ have some common prime divisor $p$.
$p \mid x \Rightarrow p \mid x^2 \Rightarrow p \mid (u - v)(u + v)$. (Eq.1)
Note that $p$ cannot be 2 since $x$ is odd. Thus $p \mid y \Rightarrow p \mid 2uv \Rightarrow uv$
$\Rightarrow p \mid u$ or $p \mid v$ (Eq. 2)
Since $u, v$ are coprime, we may assume without loss of generality in (Eq. 2) that $p \mid u$. But this contradicts (Eq. 1) since both $(u-v)$ and $(u+v)$ must be coprime w.r.t. $u$ and hence neither of them can have $p$ as their divisor. Thus the assumption of there being a common prime divisor for $x,y$ must be wrong and hence the triplet must be primitive.

Proof for the only if part:
Suppose $(x, y, z)$ is a primitive Pythagorean triplet with $x^2 + y^2 = z^2$. We already proved that any such triplet must have an odd $z$ and exactly one odd number on LHS. Without loss of generality, let us assume that $x$ is the odd number on LHS (and so $y$ must be even). Since both $x$ and $z$ are odd and $z > x$, we may rewrite them as $x = (a - b), z = (a+b)$ where $a, b$ are positive integers given by $a = \dfrac {x + z} {2}$ and $b = \dfrac {z - x} {2}$. (Eq. 3)

Thus $y^2 = z^2 - x^2 \Rightarrow (a+b)^2 - (a-b)^2 = 4ab \Rightarrow y = 2 \sqrt {ab}$. (Eq. 4)
Since $y$ is an integer, $ab$ must be a perfect square. (Eq. 5)

$a, b$ must be coprime since otherwise $ab$, $(a-b)$ and $(a+b)$ will have some common prime divisor $p$ which would contradict the initial assumption that $(x, y, z)$ have no common divisor. This condition together with (Eq. 5) implies that $a$ and $b$ must themselves be perfect squares coprime w.r.t. each other. Thus we may write $a=u^2, b=v^2$ for some coprime integers $u, v$ with $u > v$. Substituting for $a, b$ in (Eq. 3) and (Eq. 4) we get $x = u^2 - v^2, z = u^2 + v^2$ and $y = 2 \sqrt {u^2 v^2} = 2uv$. Hence proved.

 

[fresh_42]

"Question 12"
First we prove by contradiction that for any primitive Pythagorean triplet $(x, y, z)$ ($z^2$ being the sum on RHS), exactly one of either $x, y$ must be odd, while the other must be even.

• If both $x, y$ are even, then the RHS $z^2$ would also be even, in turn implying that $z$ is also even. This in turn implies that 2 is a common divisor of $x, y, z$, contradicting the requirement of primitiveness.
• If both of $x, y$ are odd, they can be written as $x=(4a + r_1), y=(4b + r_2)$ for some positive integers $a, b$ such that $r_1 = x \mod 4, r_2 = y \mod 4$ and $r_1, r_2 \in \{1, 3\}$. Then the Pythagorean sum becomes $z^2 = 4a^2 + 8ar_1 + r_{1}^{2} + 4b^2 + 8br_2 + r_{2}^{2} \\ \Rightarrow z^2 \mod{4} \equiv (r_1^2 + r_2^2) \mod{4} \equiv 2 \mod{4}$. This implies that $z^2$ is even but not a multiple of 4. But this would mean that there can be no integer solution for $z$ since an odd $z$ cannot given an even $^2$ and an even $z$ will mean $z^2 \equiv 0 \mod{4}$, again a contradiction.

Hence the only possibility is that one of $x,y$ is odd and the other is even. This implies $z^ = x^2 + y^2$ is odd and hence $z$ too must be odd.

(Note to self:) This was only shown for primitive triplets. $x\equiv y\equiv 0 \;(2)$ isn't ruled out, yet.

Proof for the if part:
Given $u, v$ are coprime with $u > v$ and exactly one of them is odd. Let $x = u^2 - v^2, y = 2uv, z = u^2 + v^2$. $x^2 + y^2 = (u^4 + v^4 - 2u^2v^2) + 4u^2v^2 = (u^4 + v^4 + 2u^2v^2) = (u^2 + v^2)^2$
$\Rightarrow x^2 + y^2 = z^2$. Thus $x, y, z$ form a Pythagorean triplet. To prove that it is a primitive triplet, let us assume that on the contrary $x, y$ have some common prime divisor $p$.
$p \mid x \Rightarrow p \mid x^2 \Rightarrow p \mid (u - v)(u + v)$. (Eq.1)
Note that $p$ cannot be 2 since $x$ is odd.

If I got you right, then you want to show that the triplet is primitive. But $x\equiv y\equiv 0 \;(2)$ isn't ruled out, yet. Only for primitive triplets, which we don't know it is, yet. Hence why is $x$ odd?

Thus $p \mid y \Rightarrow p \mid 2uv \Rightarrow uv$
$\Rightarrow p \mid u$ or $p \mid v$ (Eq. 2)
Since $u, v$ are coprime, we may assume without loss of generality in (Eq. 2) that $p \mid u$. But this contradicts (Eq. 1) since both $(u-v)$ and $(u+v)$ must be coprime w.r.t. $u$ and hence neither of them can have $p$ as their divisor. Thus the assumption of there being a common prime divisor for $x,y$ must be wrong and hence the triplet must be primitive.

Proof for the only if part:
Suppose $(x, y, z)$ is a primitive Pythagorean triplet with $x^2 + y^2 = z^2$. We already proved that any such triplet must have an odd $z$ and exactly one odd number on LHS. Without loss of generality, let us assume that $x$ is the odd number on LHS (and so $y$ must be even). Since both $x$ and $z$ are odd and $z > x$, we may rewrite them as $x = (a - b), z = (a+b)$ where $a, b$ are positive integers given by $a = \dfrac {x + z} {2}$ and $b = \dfrac {z - x} {2}$. (Eq. 3)

We only have $z^2>x^2$. If you write $z>x$ then you already made an assumption. It can be made since we can restrict ourselves to natural numbers, but why?

Thus $y^2 = z^2 - x^2 \Rightarrow (a+b)^2 - (a-b)^2 = 4ab \Rightarrow y = 2 \sqrt {ab}$. (Eq. 4)
Since $y$ is an integer, $ab$ must be a perfect square. (Eq. 5)

$a, b$ must be coprime since otherwise $ab$, $(a-b)$ and $(a+b)$ will have some common prime divisor $p$ which would contradict the initial assumption that $(x, y, z)$ have no common divisor. This condition together with (Eq. 5) implies that $a$ and $b$ must themselves be perfect squares coprime w.r.t. each other. Thus we may write $a=u^2, b=v^2$ for some coprime integers $u, v$ with $u > v$. Substituting for $a, b$ in (Eq. 3) and (Eq. 4) we get $x = u^2 - v^2, z = u^2 + v^2$ and $y = 2 \sqrt {u^2 v^2} = 2uv$. Hence proved.

Why can we write all triplets this way? What about those which are not primitive?

 

[mathwonk]

SPOILER: #9:In reference to fresh42's post #88, yes, I had in mind looking at a map f from an interval [0,1] to X that sends both end points to the same base point p lying in UmeetV. Then subdivide the interval until each subinterval goes either entirely into U or entirely into V, (or both). I will assume there are only two subintervals, [0, 1/2] and [1/2, 1], with the first subinterval mapping entirely into U and the second one mapping entirely into V, (since you just repeat the argument if there are more). Then since the point f(1/2) = q lies in both U and V, we can find a path connecting q to p and lying in UmeetV. By adjoining this path to the map f on [0,1/2], we get a loop based at p, and lying entirely in U. Then by also adjoining this path in the other direction (from p to q) to the map f on [1/2,1] we get another loop based at p and lying entirely in V. Moreover if we adjoin these two extended loops end to end, the two path components joining q to p and then back again, cancel out, leaving us with our original f. Thus the new loop, which is a product of one loop in U and one loop in V, is homotopic to the original loop f, rel base points. Since both U and V have trivial fundamental group, both factors of the new product loop are homotopic to a constant, hence so is their product and so is the original f. This is what I called the surjectivity part of Seifert Van Kampen, and thanks to you I now realize it is rather elementary!

 

[Infrared]

Right, the surjectivity in Van Kampen isn't so mysterious! Even easier than surjectivity is the fact that the subgroup identifying classes in $\pi_1(U)$ and $\pi_1(V)$ coming from the same element of $\pi_1(U\cap V)$ is contained in the kernel. The meat of the theorem is that nothing else lies in the kernel. Hatcher's argument looks pretty direct but also un-fun to read (though I guess I should sit myself down and go through it since I don't remember it exactly...)

 

[mathwonk]

I didn't read Hatcher's argument either, but was enlightended by his statement, emphasizing the surjectivity as the first step. I just visualized this argument in my head afterwards. nice point about the inclusion of the natural candidate subgroup in the kernel.

 

[Office_Shredder]

I allegedly took a class on algebraic topology in 2011, but this question and reading material about your posts was very educational.

 

[graphking]

No one try Q5？ I believe it requires the Gauss Law of Quaduatic Reciprocity, which is a very beautiful formula in general number theory, related to the equation with variable x: x^2=a(modb).

 

[Not anonymous]

But $x \equiv y \equiv 0\ (2)$ isn't ruled out, yet. Only for primitive triplets, which we don't know it is, yet. Hence why is $x$ odd?

In the if-part of the proof the initial conditions stated not only that are $u, v$ are coprime but also that exactly one of them is odd and the other one is even. Since the square of an odd number is odd and the square of an even number is even, it follows that exactly one of $u^{2}, v^{2}$ must be odd and the other must be even. $x = u^{2} - v^{2}$ in that section of the proof and we know that the difference between an odd number and an even number (regardless of whether odd is subtracted from even or vice-versa) must be odd, hence $x$ must be odd.

We only have $z^{2} > x^{2}$. If you write $z > x$ then you already made an assumption. It can be made since we can restrict ourselves to natural numbers, but why?

Sorry, I don't know what assumption I have made other than $x, y, z$ being natural numbers, and even that assumption follows directly from the definition (as I know of it) of Pythagorean triples, since unless otherwise stated, such triples must consist only of positive integers. Should it be proven that the squares of natural numbers also follow the same ordering as the natural numbers (i.e. $a > b \Leftrightarrow a^{2} > b^{2}$ for natural numbers $a,b$), or are you referring to some other assumption that I am unable to identify?

Why can we write all triplets this way? What about those which are not primitive?

If $(x, y, z)$ form a non-primitive Pythagorean triple, then by definition they have some common divisor greater than 1. Let $p > 1$ be the greatest common divisor of $x, y$. Then we can write $x = px_{0}, y = py_{0}$ for some natural numbers $x_0, y_0$ that would be coprime. $x^2 + y^2 = z^2 \Rightarrow p^2 x_{0}^{2} + p^2 y_{0}^{2} = z^{2} \Rightarrow p^2 (x_{0}^{2} + y_{0}^{2}) = z^2$. Since $p^2 | z^2$ and $z^2$ is a perfect square, $z^2$ can be written as $p^2 z_{0}^{2}$ for some natural number $z_{0}$. Dividing the Pythagorean triple equation by $p^{2}$ gives a different Pythagorean triple $x_{0}^{2} + y_{0}^{2} = z_{0}^{2}$ which must be primitive since $x_{0}, y_{0}, z_{0}$ cannot have a common divisor greater than one since $x_{0}$ and $y_{0}$ are coprime. Thus any non-primitive Pythagorean triple can be reduced to a primitive Pythagorean triple and can be generated by multiplying a primitive Pythagorean triple by some positive integer greater than 1.

 

[fresh_42]

The only-if part looks correct.

By assuming $u>v$ you lost your advantage to interchange the two. Without any need, by the way, because the assumption that we haven't all integers is artificial as it is unnecessary. We only consider squares, so $x,y,z$ can as well be negative.

For the if-part: Why is $x$ odd?
($x^2=u^2-v^2=(2k)^2-(2l+1)^2\equiv 1\vee 3 \mod 4\Longrightarrow x\not\equiv 0\mod 2\quad \checkmark$.)

Ok. Now $p\,|\,u$ or $p\,|\,v$ and $p\neq 2.$ As mentioned, you cannot assume $p\,|\,u$ because you already demanded $u>v.$ However, I assume that neither assumption has to be made. Let's see.
($p\,|\,u \wedge p\,|\,x\Longrightarrow x^2=p^2k^2=(u^2-v^2)=(p^2m^2-v^2)\Longrightarrow p\,|\,p^2\,|\,v^2 \Longrightarrow p\,|\,v.$ The same argument can be used if $p\,|\,v.$)

Thus $x$ and $y$ are coprime.

Ok. I'm convinced now. That leaves only the part to be proven, namely the statement that all Pythagorean triples can be found that way, primitive or not.

 

[Not anonymous]

By assuming $u > v$ you lost your advantage to interchange the two

$u > v$ is a condition already mentioned in the question, so it is not an assumption introduced in the proof.

That leaves only the part to be proven, namely the statement that all Pythagorean triples can be found that way, primitive or not

Sorry, I am confused again. The original question refers to only primitive Pythagorean triples and says "and which are primitive (no common divisor of $x, y, z$)", so the proof I attempted to provide was also specific to primitive triples. The proof does not say that non-primitive Pythagorean triples too can be expressed as $(u^2 - v^2, 2uv, u^2+v^2)$ for coprime $u, v$. However, since your previous reply asked about non-primitive triples, my reply to that said that any non-primitive triple can be derived from a primitive triple by multiplying every element of the the primitive triple by the same positive integer. So if $(x, y, z)$ is a primitive Pythagorean triple, then $(px, py, pz)$ will also be a Pythagorean triple for any natural number $p$, but it will be non-primitive when $p > 1$.

As an example of a non-primitive Pythagorean triple, we take $(9, 12, 15)$. It is derived from the primitive Pythagorean triple $(3, 4, 5)$ using the multiplier 3. And we cannot express $9 = u^ - v^2, 6 = 2uv$ for any coprime $u, v$ since if they were coprime, then $u^2 - v^2$ and $2uv$ will also be coprime whereas 9 and 6 are not coprime.

 

[fresh_42]

statement one:

Show that all Pythagorean triples $x^2+y^2=z^2$ can be found by
$(x,y,z)=(u^2−v^2,2uv,u^2+v^2)$ with $u,v\in \mathbb{N}\, , \,u>v$

statement two:

and which are primitive (no common divisor of $x,y,z$) if and only if $u\, , \,v$ are coprime and one is odd and the other one even.

$u>v$ is an add-on at the end, it is not really required. If we find such a representation, then it can be assumed. Anyway, the signs are not of importance.

 

[Not anonymous]

statement one:
Show that all Pythagorean triples $x^2 + y^2 = z^2$ can be found by
$(x, y , z) = (u^2 - v^2, 2uv, u^2 + v^2)$ with $u, v \in \mathbb{N} , u > v$

Sorry again, I am lost. I was able to find a way by which any primitive Pythagorean triple can be expressed as $(u^2 - v^2, 2uv, u^2 + v^2)$ with $u, v \in \mathbb{N}$ and that part of the proof was accepted, but I am unable to understand how and why all non-primitive triples should be expressible the same way. I take the same example of non-primitive Pythagorean triple as in an earlier reply, $(9, 12, 15)$. Since 12 is the only even number in this triple, only it can be expressed as $2uv$ for some natural numbers $u,v$. That leaves 9 to be expressed as $u^2 - v^2$. The possible solutions for $12 = 2uv$ are $(u=1, v=6)$, $(u=2, v=3)$, $(u=3, v=2)$, $(u=6, v=1)$. In none of these solutions do I get $u^2 - v^2 = 9$. In other words, $(9, 12, 15)$ appears to be a counterexample for $(x, y , z) = (u^2 - v^2, 2uv, u^2 + v^2)$. What am I missing?

 

[fresh_42]

Sorry again, I am lost. I was able to find a way by which any primitive Pythagorean triple can be expressed as $(u^2 - v^2, 2uv, u^2 + v^2)$ with $u, v \in \mathbb{N}$ and that part of the proof was accepted, but I am unable to understand how and why all non-primitive triples should be expressible the same way. I take the same example of non-primitive Pythagorean triple as in an earlier reply, $(9, 12, 15)$. Since 12 is the only even number in this triple, only it can be expressed as $2uv$ for some natural numbers $u,v$. That leaves 9 to be expressed as $u^2 - v^2$. The possible solutions for $12 = 2uv$ are $(u=1, v=6)$, $(u=2, v=3)$, $(u=3, v=2)$, $(u=6, v=1)$. In none of these solutions do I get $u^2 - v^2 = 9$. In other words, $(9, 12, 15)$ appears to be a counterexample for $(x, y , z) = (u^2 - v^2, 2uv, u^2 + v^2)$. What am I missing?

You are right. I made a mistake (sloppy translation). It should have been ... of the form $d\cdot (u^2-v^2,2uv,u^2+v^2)$ ...

I apologize for that negligence, @Not anonymous.

 

[Not anonymous]

I apologize for that negligence, @Not anonymous.

No problem (problem ≠ math problem in this context ). Thanks to you as always for promptly reviewing my answers and patiently explaining what I may have missed or got wrong.