@Huy You know, I can answer questions too. :P Yes: there's a covering map homomorphism $S^3 \to SO(3)$, and covering maps induce an isomorphism on the Lie algebra.
If not, I guess the answer is just "yes, you can write down an isomorphism between the two". But actually the reason you can is that they have the same universal cover: simply connected Lie groups are determined up to isomorphism by their Lie algebra, and the Lie algebra is preserved under passing to the universal cover. Because $S^3$ (as a group!) is the universal cover of $SO(3)$, they have the same Lie algebra.
If one differentiates the map $SU(2)$ to $SO(3)$ where one sends a unit quaternion $q$ to the map span(i,j,k) to span(i,j,k) given by conjugating by $q$, you can see explicitly it is an isomorphism of Lie algebras. This is done somewhere on this site.
And I guess I suggested to an undergraduate that he should do it in his introductory talk on Lie groups.
@PVAL One version of the Malcev-Iwasawa theorem says that any Lie group $G$ has a maximal compact subgroup $K$, and $G$ is diffeomorphic to $K \times \mathbb R^n$. An immediate corollary is that $G$ cannot possibly be an exotic $\mathbb R^4$. Is there a more elementary way of proving this? (For instance, is the exponential map a diffeomorphism?)
@MikeMiller: In some lecture notes I found online, there's a rather short proof that $RP^n$ is orientable iff $n$ is odd. The proof that it is not orientable when $n$ is even goes as follows: Let $\pi: S^n \to RP^n, p \mapsto [p]$. Let $r: S^n \to S^n$ be reflection through the origin. Then, $\pi \circ r = \pi$. If $RP^n$ is orientable, then we may assume that $\pi$ preserves orientation (why?). From this follows that $r$ preserves orientation as well.
They claim that this is only the case when $n$ is odd. Shouldn't this be the other way around, since $(-1)^n = 1$ iff $n$ is even?
@Huy: If $M$ is an oriented manifold and $G$ is a discrete group of diffeomorphisms of $M$ (acting freely), $M/G$ is oriented iff $G$ acts by orientation-preserving diffeomorphisms. The formula for the degree of the antipodal map is $(-1)^{n+1}$, not $(-1)^n$. (It's the composition of $n+1$ reflections.)
In any case if you have a covering map $\pi: M \to N$ between oriented manifolds, it either preserves or reverses the orientation. If it reverses the orientation, well, just change one of the orientations involved.
@MikeMiller Probably a very stupid question : $S^1 \to M$ be an embedding into a, say, 3-manifold $M$. If this thing is homotopic to a point, is it necessarily isotopic to it?
If the dinension of the codomain is at least 5 any embedded null-homotopic circle bounds a disc by transversality theory. If the coodomain is $S^4$ it bounds a disc by Whitney's strong embedding theorem. For a general 4-manifold it's not obvious to me it has to bound. You'd want to isotope it into a single coordinate patch.
@Mike Miller I think R^4 (as a topological space) admits a unique topological group structure (I'm not sure how to do this) up to isomorphism. Any continuous isomorphism of Lie groups is smooth so that should be enough.
@Mike Miller A nullhomotopic circle in a smooth 4-manifold (without boundary) necessairly bounds a framed immersed disk $D$ with isolated self-intersections. There is an annulus $A$ diffeomorphic to $\Bbb R \times S^1$ which intersects such a framed disk in exactly the circle. There is a ribbon move which removes a self intersection of $D$ and adds in two cancelling intersection points. If one does this procedure to each intersection point, you get a framed embedded disk.
@MikeMiller It seems like one could make my question a bit more complicated by making some assumption on the disk by which one nullhomotopes the embedded circle. I dunno.
@PVAL: Is this basically just the same argument Whitney gives for $\mathbb R^4$, or do you have to make serious modifications? I never looked carefully at the proof of the strong embedding theorem.
The ribbon move is exploiting that the boundary of the circle isnt in the boundary of the 4-manifold by pushing intersections off an arc between the boundary of the disk and the intersection point. There is a better description than I can give in Freedman-Quinn. This doesn't work if the circle is in the boundary by the existence of knots that are not slice.
@PVAL When doing strong Whitney in higher dimensions, you start with a map and modify it to become an embedding. Is the modification usually not homotopic to the original?
@Mike Miller I'm not sure what you are asking. Strong Whitney usually implies maps to R^n, so all maps are homotopic. If you start with an immersion though, the map may generally not be homotopic through immersions to an embedding (i.e. if there is an odd number of self-intersections they cant appear in Whitney disks because they cant be paired up. In that case you have to add cusps.
I ask because there's a paper yesterday on the arXiv that proves an h-principle for Engel structures, which are closely related to contact structures, so maybe you'd care. I haven't read it yet but intend to skim it today.
@PVAL: Very poorly phrased on my part. I actually care about the case where you have boundary and your immersion is an embedding near the boundary. Then I want to know if the modified map is homotopic to the original held constant near the boundary.
i'm probably free associating in the wrong way, but when i see symplectic manifolds + lagrangian submanifolds my brain goes to quantization of classical mechanics
which would be amusing if there was a link, since then a cute name for Bohr-Sommerfeld quantization would be the $\hbar$-principle :P
In a simply-connected manifold $M^{2n}, n \geq 3$ , for an immersion with self-intersections there is circle between any two self-intersection points of opposite sign on the image which bounds a whitney disk. One can then resolve this disk in the obvious way in a neighborhood of this disk. This does not change the regular homotopy class of the immersion. If there aren't the correct number of self-intersection points of each sign one has to add cusps.
This changes the regular homotopy class, but does not change the homotopy class of the maps. and a cusp can be added for each missing intersection point. Adding a cusp is a local operation and can be chosen to not hit a nbhd of the boundary if you wish. The $n=1$ case is actually pretty insightful for how to do this.
I have a program that tells me the angle in radiants of a cursor from another item, let's say it's a star.
See this picture -> i.stack.imgur.com/1MoTq.png
What I want is to take this X2,Y2 point and deviate it a bit on the left or right from the cursor angle.
See this picture -> i.stack.imgur....
@DanielFischer Lars Ahlfors states, "In the region obtained by omitting these half-lines, the principal branch of $\log(1-e^{2iz})$ is hence singled-valued and analytic." The half-lines he's referring to is where $1-e^{2iz}$ is real and negative. Is this a misstatement? Wouldn't it be more appropriate to say something like, "Using the principal branch of the logarithm, $\log(1-e^{2iz})$ is single-valued and analytic in the region obtained by omitting these half-lines."
from mathematica, i'm getting $$_3\tilde{F}_2(1/2,1/2,1;n+1,n-1|x)=\left(\frac{x}{16}\right)^n \binom{2n}{n}{_2F_1}(n+1/2,n+1/2;2n+1|x)$$ for nonnegative integer $n$
where $_pF_q$ and $_p\tilde{F}_q$ are the (generalized) hypergeometric and regularized hypergeometric functions
You have the starting point and the cursor is the second point. You have the angle between those two points. You want to add 0.5 radiant to the angle and calculate the resulting "second point". The distance from the starting point remains the same
oh, @ted. in connection with the wedge product stuff we were discussing, i noticed that wikipedia mentions the following (rather elementary, i suppose) isomorphism: $\Lambda(V\oplus W)\cong\Lambda(V)\otimes \Lambda(W)$
ultimately i want to write out elements like $h^*h$ where $h\in \Lambda(H)$ and $h^*$ is its dual (in terms of row/column vectors, i want an outer product)
btw, i don't think i ever explained why i was interested in this
basically, i was trying to figure out the manipulations leading to equation (9) of this paper, but working in the language of exterior algebra rather than the usual physicist bra-ket notation
in the hopes of developing my facility with the former
I don't know how long I have. My computer seems to be having a problem. It will only hold a network connection for a short while before shutting down all networking. >8(
2
I am at the Apple Store waiting for the computer to fail
