Ok so homotopic to a point implies existence of a map from $D^2$ to the space such that the restriction of this map to the boundary of D2 gets mapped to the loop. Now we need to convert this map into a chain?
I am trying to write down how the complexification of the 4d irreducible real representation decomposes into two copies of the 2d irreducible complex representation. I was given a hint to try to find an isomorphism of vector spaces, but I don't really know what I'm doing.
But if you stay with lin.alg. you'll learn that the space of tuples of real numbers is just one example of vector spaces
But it turns out any finite-dimensional vector space with dimension $n$ is isomorphic to $\mathbb R^n$ or $\mathbb C^n$ so once you know Euclidean space very well it very naturally carries over to all finite dimensional spaces
I don't know if you came across the idea of isomorphisms yet but I can explain what that means if you want
My topology professor told us about a professor of his when that was always making sign errors, so he always used $\mathbb Z/2\mathbb Z$ to compute homologies because then $-1 = 1$
@mercio Oh okay, could you also show me how $\int_{\mathbb R}e^{\pi x_1^2}e^{-2\pi ix_1 \cdot \xi_1}$ results $e^{\pi \xi_1^2}$ ? I don't know how to approach to it when there's a dot product
How does one argue that there exist points $y_0,...,y_n \in [a,b]$ such that $y_0=a$, $y_n = b$, and $y_k - y_{k-1} < \epsilon$ for $k=1,...,n-1$? I tried using compactness, but I couldn't see how this would work.
@user193319 I don't know the context your question is in but this looks like a case for the Archimedian principle which ensures that for every $\varepsilon \gt 0$ there is an $n \in \Bbb N$ such that $\frac{1}{n} \lt \varepsilon$.
then just split your interval equally spaced parts with length $1/n$
this doesn't use compactness but rather completeness
Can somebody explain to me the difference between having a curve given parametrically as the image of a function $g(t)$ or as the graph $y = f(x)$ of some function $f$? To me they seem more or less like the same thing? I understand the difference of a curve being given implicitly by an equation. Is there a difference in the explicit representation by a parametrization or the graph of some function $f(x)$?
@LeylaAlkan there any many proofs of that integral but the one about changing to polar coordinates is so beautiful, Google it if you haven't seen it before, it's short
I think there's actually a post on math.se which is entirely proofs of the Gaussian integral
@TedShifrin Was just reading your book and I don't dare :P My question is, why do we want the curve given as the graph rather than the image of some function? Why is this better?
Ted. this has been bothering me since last night --
Suppose M is compact, 2-dimensional, oriented, embedded submanifold of R^3\{(0, 0, 0)}. Let omega := x . *dx (2-form). What are the possible values of the integral of omega over M?
So for the cycloid then we are done since we have the parametric representation and don't need a graph? My question is directed to understanding the motivation of the implicit function theorem. I watched the lectures and also the definition of manifolds, but I want to understand the motivation of all of it.
@TedShifrin Do I understand this right? The general motivation of the implicit function theorem is to be able to express implicit sets (for example curves) explicitly as the graph of some function (even if only locally) because this is better then implicitly. This has no connection to a parametric representation which would also be explicit (?).
implicit function theorems gives you local conditions for functions but doesn't tell you anything about how you can explicitly write down a formula, usually you can't
@TedShifrin Sure. Just the book reads like "we know the parametrization of the cycloid but not the function of which it is a graph". Bascially my question is: Why would we care? We know the parametrization, so isn't this good enough?
So it fails the obvious "implicit definition" — $f(x,y) = x^3-y^2 = 0$, but the derivative fails to have rank $1$ at the origin. Does that mean it can't be a manifold?
How to you prove there cannot be a chart around the origin?
That doesn't prove it is not a manifold, @GFauxPas. It's a sufficient condition, not a necessary one. Consider, for example $f(x)=x^2$, $f\colon\Bbb R\to\Bbb R$.
Isn't it because the curve is like an X around the origin so there is no way for it to be the graph for a function? Vertical/Horizontal line test fails.
Ted doesn't seem mean enough to win in a fist fight...but a guy who wages war against determinants and also does functional analysis for a living must be hard
For a second course in linear algebra, understanding the minimal polynomial is not that bad an idea. But students should see determinants and the characteristic polynomial explicitly in a first linear algebra course.
I would not tell someone to read his book as a first exposure to linear algebra. For one thing, there's hardly any geometry at all in there.
His book title shows this too. I expected "more" when I finally read the book. It was written like any other, didn't appeal more on geometric intuition.
Hi, I would like to know: What is the upper bound for number of non-leaf nodes in tree? or How many non-leaf nodes in a tree? Is there any theory about this
@Ted I think I got confused between the parametrization $g(t)$ and the graph of some function $y = f(x)$. In general does $g$ need to be a function? Because when it traces out for example a circle it looks like it fails the horizontal/vertical line test. Instead $f$ is required to be a function by the IFT, right?
Got it. I need to disconnect the parametric representation of the circle from the graph of the function the IFT guarantees to exists, because its graph lives in $\Bbb R^3$ and not $\Bbb R^2$.
