Jason DeVito
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 Apr 29 answered Question Regarding Proof of Taylor Remainder Theorem in Tu's “An Introduction to Manifolds” Apr 26 awarded Nice Answer Apr 21 comment What exactly is an embedding in differential geometry? As studiousus says, $1$ is weaker without some extra hypothesis. Here's an explicit example: Map $\mathbb{R}$ into $\mathbb{R}^2$ with image the letter $P$. Then neighborhoods of the triple point don't look like neghborhoods in $\mathbb{R}$, but the mapping is a smooth injective immersion. Apr 20 answered Even dimensional Lens spaces Apr 18 comment Orientability of Grassmannians @Ted: I'm not familiar with those forms (at least by that name). Are those the bi-invariant differential forms on $SO(n)$? And thanks for fixing my typos! Apr 18 comment determinant of SU(3) matrix $SU(3)$ is a a real Lie group whole elements are complex matrices satisfying $A\overline{A}^t = \overline{A}^t A = I$ and $\det (A) = 1$. I agree that it's 8 dimensional as a manifold. But maybe I'm misunderstanding your objection. Apr 18 comment determinant of SU(3) matrix But $SU(3)$ consists of complex matrices... Apr 13 answered Let $G$ be a Lie group and $H$ be a closed subgroup of $G$. Show that if $H$ and homogeneous space $\frac{G}{H}$ are connected, then $G$ is connected. Apr 10 comment Normal bundle associated to $\mathbb{R}P^n\hookrightarrow \mathbb{R}P^{n+1}$ is the tautological line bundle Take your time - I'm ok if you never accept this answer. (I'd also like to see a more geometric answer.) Incidentally, here's another approach math.stackexchange.com/questions/1277823/…. Apr 9 comment Group acting on a Projective Space What is the connection between the $G$ action and the basis of $V$? How is $G$ acting - presumably it's not just by random permutations on $P(V)$? Apr 9 answered Normal bundle associated to $\mathbb{R}P^n\hookrightarrow \mathbb{R}P^{n+1}$ is the tautological line bundle Apr 6 comment Area of a circle $\pi r^2$ I agree that the area of the circle of radius $r$ is larger than the area of a square with side length $r$, but it some ways, this is misleading. The point is that the $r$ on a circle only measures half a distance, while the $r$ in your square example measures a full distance. A better comparison is that a circle of radius $r$ has less area than a square of side length $2r$. In, fact, drawing a picture, you can see that the largest circle in a square fills up roughly $3/4$ of the area. (In fact, it fills up exactly $\pi/4$ of the area.) Apr 6 answered If $f(x)$ is discontinuous at $0$, does that mean ${f(x)}^3$ is discontinuous at $0$? Apr 5 comment Property of a polynomial with no positive real roots Nice! Thank you. Apr 5 comment Complex Analysis: Liouville's theorem Proof @JessyCat: The $ML$ inequality applies to $\left|\int_{|z|=R} \frac{f(z)}{(z-a)(z-b)}\; dz\right|$, not to $\int_{|z|=R} \frac{f(z)}{(z-a)(z-b)}\; dz$. (In fact, without the absolute value bars, the integral is (potentially) a complex number. What does it mean for a complex number to be $\leq 0$?) Apr 5 comment Property of a polynomial with no positive real roots How do you show $PQ\in S$ implies $P$ and $Q$ are in $S$? Apr 4 comment Are there countably many closed manifolds in each dimension? The comments to this question math.stackexchange.com/questions/109881/… imply that there are only countably many closed manifolds in each dimension. Perhaps you could distill them into one coherent response? Apr 3 comment Unnecessary assumption in exercise (from Spivak, Calculus on Manifolds) Your solution looks fine to me. Apr 3 comment For a matrix $A$, does $A^3=I_n$ mean that $\det(A)=1$? As Spenser's answer shows, it depends on the field $\mathbb{K}$. If $1$ is the only cube root of $1$ in $\mathbb{K}$, then the answer is yes. Apr 1 comment How to interpret the cotangent bundle of a complex manifold? That just shows my complex analysis is rusty - I should have known that. Thanks!