@BalarkaSen Earlier you mentioned that $\pi_6S^3$ has a $\mathbb{Z}_3$ component. Just wanted to point out that I was reading through Lecture Notes in Algebraic Topology by Kirk and Davis just now and in Corollary 10.13 it's stated that "If $p$ is a prime, the $p$-primary component of $\pi_iS^3$ is zero if $3 < i < 2p$ and is $\mathbb{Z}_p$ if $i = 2p$" which seems to be a stronger result
Let's see. Take the generator $S^3 \to K(\Bbb Z, 3)$ of $\pi_3 K(\Bbb Z, 3)$ and let $F$ be the homotopy fiber, then running the homotopy LES on $F \to S^3 \to K(\Bbb Z, 3)$ gives $\pi_i F \cong \pi_i S^3$ for $i > 3$ and $\pi_i F = 0$ for $i \leq 3$. Take the homotopy fiber of $F \to S^3$ in turn, let's call it $F'$. Then running the homotopy LES on $F' \to F \to S^3$ we see that $\pi_i F' = 0$ for $i > 3$, $\pi_3 F' = 0$, $\pi_2 F' = \Bbb Z$ and $\pi_1 F' = 0$. So $F' = K(\Bbb Z, 2)$.
So run the Serre spectral sequence on $K(\Bbb Z, 2) \to F \to S^3$.
$E^2_{p, q} = H^p(S^3; H^q(K(\Bbb Z, 2))$, which is nonzero only for $p = 0$ and $p = 3$. Just two of those columns full of $H^* K(\Bbb Z, 2)$.
No nonzero differential at the $E^2$ page but there's a good differential at the $E^3$ page, namely $d^3 : E^3_{0, 2} \to E^3_{3, 0}$ and after that no nonzero differential comes out/goes in from those positions, so this guy has to be an isomorphism otherwise $E^\infty$ page will have some nonzero term in $(0, 2)$ or $(3, 0)$ but that'd force $H^2 F$ or $H^3 F$ to be nonzero, which is not possible by Hurewicz + $\pi_2 F = \pi_3 F = 0$.
This is a differential $H^0(S^3; H^2(K(\Bbb Z, 2)) \to H^3(S^3; H^0(K(\Bbb Z, 2))$. Explicitly write the cohomology rings $H^* S^3 = \Bbb Z[x]/(x^2)$ where $|x| = 3$ and $H^* K(\Bbb Z, 2) = \Bbb Z[y]$ where $|y| = 2$. So $d^3(1 \otimes y) = x \otimes 1$. The strategy is to use the product structure in the spectral sequence now
$d^3 : H^0(S^3; H^{2k}(K(\Bbb Z, 2)) \to H^3(S^3; H^{2(k-1)}(K(\Bbb Z, 2))$ is given by $d^3(1 \otimes y^k) = k x \otimes y^{k-1}$ by Leibniz rule. So at the group level $d^3 : E^3_{0, 2k} \to E^3_{3, 2(k-1)}$ is multiplication by $k$.
We know the kernel and cokernel of that! So the spectral sequence degenerates at $E^4$ and in the $E^\infty$ page we have $\Bbb Z/k\Bbb Z$ arranged along the even terms in $3$rd row and nothing else. So $H^{2k+1} F = \Bbb Z/k\Bbb Z$ and $H^{2k} F = 0$.
Go back to homology so that it reads $H_{2k} F = \Bbb Z/k\Bbb Z$ and $H_n F = 0$ if $n$ is odd. Now to compute homotopy groups of $S^3$ out of this...
By induction assume the $p$-primary component of $\pi_i S^3$ is zero for $3 < i \leq k < 2p$ (this is easily checked with $i = 4$ since $\pi_4 S^3 = \Bbb Z_2$ and by hypothesis $p$ is at least $3$). Now $\pi_i S^3 = \pi_i F$, so by Hurewicz theorem mod C (where C is the class of p-groups), $\pi_{i+1} F \cong H_{i+1} F$ mod C.
If $i < 2p$, then $H_{i+1} F$ is either $0$ or $\Bbb Z/n \Bbb Z$ for some $n < p$, so $H_{i+1} F$ has $p$-primary component zero as well. That means $\pi_{i+1} F$ has $p$-primary component zero!
Finally, $\pi_{2p} S^3 = \pi_{2p} F = H_{2p} F = \Bbb Z/p\Bbb Z$ (mod C), i.e., $\pi_{2p} F$ has $p$-primary component $\Bbb Z/p\Bbb Z$. So this proves your assertion :)
It's more for my own practice than anything haha but glad it's helpful
I suppose what one can do is instead of playing of Hurewicz mod C games at the end is localize the spaces involved at the prime ideal $(p)$ from the very onset. I don't know if there's an advantage to that though
As a math educator, do you think it is appropriate to insist that students say "negative $0.8$" and
not "minus $0.8$" to denote $-0.8$?
The so called "textbook answer" regarding this question reads:
A number and its opposite are called additive inverses of each other because their sum is zero,...
I saw another post of this exact problem , just different polynomial , and the guy with the answer proposed some translations on imaginary axis and I don't frankly know what
But it seems to be correct
Just lazy to study that unusual method 2 hours before the exam
And honestly I dont think we are meant to use that method in the exam
Given a Vector space $V$ , if $\Omega$ is a symplectic form on $V$ and $J \in Aut(V)$, I define the bilinear form $g(v , w) = \Omega(v, Jw)$. If I add the condition that $J^2 = -\mathbb{id}$ does that imply $g$ is symmetric?
I don't think it does, but my book says that $g$ is symmetric. The maximum I was able to prove was that $g( Jv , w) = g(v , Jw)$
Hello all. If I want to show that a $1$-form $\omega$ defined on $\Bbb R^4$ in Cartesian coordinates is nonvanishing when I pull it back to the $3$-sphere over the natural inclusion map, do I actually need to bother pulling it back with respect to a parametrisation
Say it's just something nice like $\alpha = xdy -ydx + z dt -t dz$, can I just observe that for no $x,y,z,t\in\Bbb R^4$ do all $4$ of $x,-y,z,-t$ vanish
Is there any easy way to prove darboux theorem? The one I have done is quite tedious. It involves taking the derivative with time of this 1-parameter family of differential forms pulled back under a diffeomorphism between two manifolds.
I don't understand what's complicated about it. Given any 1-parameter family of $k$-forms $\alpha_t$ you want a flow which takes $\alpha_1$ to $\alpha_0$. That won't be a global vector field in general, it would be a time-dependent one, let's call it $X_t$. So you want to solve for $\phi_t^* \alpha_0 = \alpha_t$ where $\phi_t$ is infinitsimally generated by $X_t$
Differentiate that to get $\mathcal{L}_{X_t} \alpha_0 = \partial_t \alpha_t$
@BalarkaSen when you type (why?) in parenthesis I think maybe I'm suppose to see why without pulling back? But when I pull back I get the zero form so it is definitely true
This is the starting point. The Moser lemma says that if $\alpha_t$ is a family of symplectic forms in a specific cohomology class, so that $\alpha_t = \alpha_0 + d\beta_t$ for some $\beta_t$, then the solution to $\mathcal{L}_{X_t} \alpha_0 = \partial_t \alpha_t$ is given by uh setting $X_t$ to be the $\alpha_t$-dual of $\partial_t \beta_t$
Typo: The formula should have been $\mathcal{L}_{X_t} \alpha_t = \partial_t \alpha_t$. My differentiation was wrong.
Or at least misleading notations
The last thing I said is a very easy check: Let $X_t$ be such that $\partial_t\beta_t = \alpha_t(X_t, -)$. Then $\mathcal{L}_{X_t} \alpha_t = \iota_{X_t} d\alpha_t + d \iota_{X_t} \alpha_t = d \partial_t \beta_t = \partial_t d\beta_t = \partial_t \alpha_t$
@user681391 You can sort of see in two ways, one is that if $p = (a, b)$ is a point on the unit circle with (unit) tangent vector $v = (-b, a)$ then if $\omega = xdx + ydy$ then $\omega(p, v) = a(-b) + b(a) = 0$.
This means $\omega$ vanishes on every tangent space to the unit circle, which is the same as saying the pullback is zero
The other is to note that $x^2 + y^2 = 1$, so just "differentiating that", $2xdx + 2ydy = 0$ :P
This is the implicit function theorem at work, if you want
This is the less tedious, without pullbacks, way to do it
So I can define a $3$-frame at every point for the tangent space at each point, given by $T_xS^3=\{v\in\Bbb R^4\mid x\perp v\}$ and check it for that for example
Is there any reason why you should expect something like darboux theorem for symplectic two forms? Some kind of geometry picture associated with it? I kind of have a feel of why symplectic geometry must be governed by global invariants rather than local because using the symplectic form you can define the volume form which gives you a definition of orientability and in a neighbourhood around a point its always orientable. I don't think this has got to do anything with locally having the same structure but this was just my thought process behind looking at darboux theorem.
If I want to write it in a 2 by 2 form then the diagonal will be 0 and the identity matrix on the top right row and negative of the identity matrix on the bottom left, if I am not wrong
That gives you the standard symplectic structure on $\Bbb{R}^{2n}$
@Albas Right, so there's always a basis $(e_1, \cdots, e_n, f_1, \cdots, f_n)$ on the vector space such that the billinear form is $e^1 \wedge f^1 + \cdots + e^n \wedge f^n$
This is the "linear Darboux theorem". A not so hard linear algebra exercise
If $\omega$ is an almost symplectic form on $M$, then for every point $p \in M$, you can find basis $(e_1, \cdots, e_n, f_1, \cdots, f_n)$ such that $\omega = \sum_i e^i \wedge f^i$. In fancy language $(e_1, \cdots, f_n)$ constitute an abstract section of the frame bundle $\text{Fr}(TM) \to M$ along which $\omega$ is the standard symplectic form.
The condition $d\omega = 0$ is an integrability condition, which says this abstract section of the frame bundle can be chosen to be locally integrable, i.e., comes as derivative of a local parametrization $\phi : U \subset \Bbb R^{2n} \to M$.
@BalarkaSen I had thought of that condition as coming from the Hamiltonian vector field(written using the Poisson bracket) and taking the lie derivative of the two form along it using the magic formula, which did give that condition.
What are some interesting things people do with symplectic geometry. I am interested in understanding Geometric Quantization but besides that? You can always try to find global invariants but besides that?
Finding global invariants is not a small deal but just asking.
@Albas OK, so you might ask, a Riemannian metric on a tangent space is just a positive definite symmetric form, so you can cook up an abstract section of the frame bundle along which the Riemannian metric is just the usual dot product. What's the local integrability condition there? That's the vanishing of the Riemann curvature tensor
But studying flat metrics is just... not that interesting. One way to think about that would be that vanishing of the Riemann curvature tensor is a second order differential restriction on the metric; on a normal coordinates the second order term in the Taylor expansion of the metric is $R$
So what's the first order restriction? Vanishing of the torsion.
The analogue of Darboux for a metric then is existence of normal coordinates
There are other examples scattered throughout geometry. If $J$ is an "almost complex structure" on $M$, i.e., $J : TM \to TM$ is a bundle automorphism such that $J^2 = -I$ fiberwise, then the integrabiliy condition for $J$ is the vanishing of the Nijenhuis tensor $J[X, Y] - [JX, Y] - [X, JY] - J[JX, JY] = 0$. If this happens, then $J$ integrates to an actual complex structure on $M$. The analogue of Darboux is existence of complex charts
That's the Newlander-Nirenberg theorem
Just so happens that for symplectic structures $d\omega = 0$ turns out to be a pretty simple integrability condition
Yeah. Hmm this does make me wonder. Having this two form imposes this cohomology condition on the manifold that $H^2(M)$ (sorry if this is the wrong notation) is not trivial. Are there similar things in cases such as the complex structure?
@Albas Even admitting an almost complex structure is a strong restriction; only spheres which admit almost complex structures are $S^2$ and $S^6$ (I think a proof involves characteristic classes). I also think there are nontrivial cohomological obstructions, since there are natural bigradings on the de Rham complex arising out of holomorphic and anti-holomorphic form which I think relates to Hodge theory?
It's pretty believable that complex structures are very rigid if you ask me, but I'd be hard pushed to give concrete examples of the topological obstructions, not knowing much about them myself
I'm trying to see why $$dx^1\wedge dx^2+dx^3\wedge dx^4+\dots + dx^{2n-1}\wedge dx^{2n}$$ is indecomposable, so I took arbitrary $1$-forms $\alpha = \sum_{i=1}^{2n} a^idx^i$ and $\beta = \sum_{i=1}^{2n}b^idx^i$ and then took $$\alpha\wedge \beta =\sum_{1\leq i<j\leq 2n} (a^ib^j-a^jb^i)dx^i\wedge dx^j$$ which gives me lots of equalities, but I can't see how to deduce that these are all impossible to satisfy
@user681391 If a 2-form $\omega = \alpha \wedge \beta$ is decomposable then $\omega \wedge \omega = 0$ (playing off of the fact that square of a 1-form is zero)
(In fact verify that if $\omega$ is your form then not only $\omega \wedge \omega \neq 0$ but $\omega^{\wedge n} \neq 0$)
$\omega^n = n! dx^1\wedge \dots\wedge dx^n$ I think, since I only need to move even position things to even position things and odd position things to odd position things, which uses an even number of transpositions in each case
Oh right, $\alpha\wedge \beta = (-1)^{deg(\alpha)deg(\beta)}\beta\wedge \alpha$ which means for odd degree $\alpha\wedge\alpha = 0$ but for even degree they commute
Was meant to say $n!dx^1\wedge\dots\wedge dx^{2n}$
Ha ha ... There is a lot more geometry than there was when I first started chatting 4 or 5 years ago.
BTW, the coordinate-free expression is in textbooks, not just on-line. And it is highly useful (e.g., for understanding/proving the Frobenius theorem).
First, check that the right-hand side defines a tensor — i.e., is linear over $C^\infty$ functions. Then it's easy to check by choosing $X=\partial/\partial x^i$ and $Y=\partial/\partial x^j$.
Except in the case of a closed hypersurface, you don't get a canonical orientation (indeed, it might not even be orientable — take the Klein bottle in $\Bbb R^4$) on a submanifold.
Why? We all know that you can parametrize a circle by $[0,2\pi)$ unofficially, and by two charts $(-\pi,\pi)$, $(\pi/2,3\pi/2)$ or whatever, officially.
Yes, but that extends continuously with no problem (unlike what happens if you remove a circle from the Klein bottle and orient what's left).
You can pull back the $2$-form by the inverse of the parametrization, or you can designate what will be the oriented basis of the tangent plane at each point.
Teaching style also changes from person to person. Like, I've had professors who just throw numbers out and professors who describe the closure axiom for groups by saying "If I add two and two, I had better not get a chicken."
though my steak lunch with the higher-ups (and the other new employees) got postponed to a later date. I was looking forward to the company paying me both for the lunch and for the time spent eating it.
So for $(u,v)\in (0,2\pi)\times (0,2\pi)$ we take $$\Phi(u,v)=(\cos u,\sin u, \cos v,\sin v)$$ and then where it's defined $\Phi^{-1}(x,y,z,w)=(\arccos(x),\arcsin(y))$
in which case $(\Phi^{-1})^*(dx^1\wedge dx^2)=d(\arccos(x))\wedge d(\arcsin(y))=-\frac{1}{\sqrt{1-x^2}}dx \wedge \frac{1}{\sqrt{1-y^2}} dy$
where I feel like I'm messing everything up somehow lol
Unless that's right
My main concern is there should be u's and v's lol
Maybe it's better to say $\Phi^{-1}(\cos(u),\sin(u),\cos(v),\sin(v))=(u,v)$ since it's bijective and then proceed as I did, in which case I end up actually having some nice $u,v$ terms and: $$-\frac{1}{\sqrt{1-u^2}\sqrt{1-v^2}} du\wedge dv$$
Nah that's bad, then $u,v=1$ destroys me
Maybe I want to choose $(u,v)\in (1,2\pi+1)^2$ or something...
I am currently looking through the codebase we are taking over from another company. One thing it needs to do get criminal records for a bunch of people to see if they are eligible for a hunting permit. When it sorts the people into those with and without a criminal record, the people who wrote the original code decided to call the lists badBoys and goodBoys.
@F.White This is messed up a bit. You want $(\arccos x,\arccos z)$, and then you just note that $dx\wedge dz$ gives a positive $2$-form. I actually think the right coordinates to use on the torus are $(u,v)$ in the first place. And, of course, $du\wedge dv$ gives the $2$-form. Why do we need ambient coordinates? That's not so natural, but it's fine.
Let $B$ be a Banach algebra and let $B^*$ denote its associated continuous dual, which can be normed in a natural way (sup. norm over unit ball in $B$). Is it true that $\phi_n \to \phi$ in $B^*$ if and only if $\phi_n \to \phi$ uniformly on the unit ball in $B$?
I don't like what he's doing with no explanation, but he's comparing the wedge product of normal $1$-form (which annihilates the sphere) and the area $2$-form. (Forget about $\iota_X(\Omega)$ for now.) Think about $(x\,dx + \dots)\wedge\Omega$. Is this $+$ the standard volume form or $-$ the standard volume form?
As a math educator, do you think it is appropriate to insist that students say "negative $0.8$" and
not "minus $0.8$" to denote $-0.8$?
The so called "textbook answer" regarding this question reads:
A number and its opposite are called additive inverses of each other because their sum is zero,...
