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Apr
20
comment Homology Whitehead theorem for non simply connected spaces
@studiosus But as I understand it, it is only required that the Euler characteristics be equal and $m=1$, and it seems to me he is saying that $m=1$ whenever $\chi>\chi_{\min}(G,2)$. Since we know by example (the complex associated to the presentation $\langle a,b\mid a^3=b^2\rangle$) that there is a $[G,2]$-complex with $\chi=0$ we have $0\geq\chi_{\min}(G,2)$, and so, for any $[G,2]$-complex with $\chi\geq 1$, $m=1$. My point is that it apparantly doesn't matter that $\chi_{\min}(G,2)=0$ or not, it only matters that it be less than our $\chi$. Am I correctly interpreting the article?
Apr
20
revised Homology Whitehead theorem for non simply connected spaces
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Apr
20
revised Homology Whitehead theorem for non simply connected spaces
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Apr
20
revised Homology Whitehead theorem for non simply connected spaces
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Apr
20
revised Homology Whitehead theorem for non simply connected spaces
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Apr
20
comment Homology Whitehead theorem for non simply connected spaces
I left an answer to tell you what I understood from your examples, could you take a look at it and correct my understanding of it?
Apr
20
answered Homology Whitehead theorem for non simply connected spaces
Apr
20
answered Is $\sin(\arcsin(x))$ equal to $x$?
Apr
20
comment Euler characteristic and free action
What action of $G$ on $K^2$ are you considering?
Apr
19
revised For two p.d. matrices $A$ and $B$, prove that $\lambda_1(AB)\leqslant \lambda_1(A) \cdot\lambda_1(B)$
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Apr
19
comment Let $A$ a matrix with real or complex entries. Proof that $\displaystyle\lim_{n\rightarrow\infty}(E+\frac{A}{n})^n=e^A, E=$indentity.
You can probably show that this sequence of functions converges uniformly on every compact set (equipping $M_n(\Bbb C)$ with some submultiplicative norm for greater convenience), so that the function $A\mapsto\lim_n\left(I+\frac{A}{n}\right)^n$ is continuous. It is then easy to show that it agrees with the usual exponential on the dense set (in $M_n(\Bbb C)$) of diagonalizable matrices, and hence the standard exponential and this agree on all of $M_n(\Bbb C)$.
Apr
19
answered For two p.d. matrices $A$ and $B$, prove that $\lambda_1(AB)\leqslant \lambda_1(A) \cdot\lambda_1(B)$
Apr
19
comment Help trying to identify a set and determine whether it is a subspace of $\Bbb{R}^n (n>2)$
Recall the definition of a subspace, and verify the axioms. You only have to prove $S$ is non empty, stable under sum and scalar multiplication.
Apr
18
comment Is $\Lambda(T^{*}E)=\bigoplus_{k=0}^n\Lambda^k(T^{*}E)$ a complex line bundle over $T^{*}E$?
It's not a vector bundle over $T^*E$! It's a vector bundle over $E$!!!
Apr
18
comment Is $\Lambda(T^{*}E)=\bigoplus_{k=0}^n\Lambda^k(T^{*}E)$ a complex line bundle over $T^{*}E$?
Usually you'd include the 0-th exterior power $\Lambda^0(T^*E)$ (which is a line bundle) in the definition of $\Lambda(T^*E)$; in any case, $\Lambda(T^*E)$ (with the correct definition) is a vector bundle of dimension $2^n$ over $E$ (not $T^*E$) where, I take it, $E$ is a manifold of (real) dimension $n$.
Apr
18
comment Is $\Lambda(T^{*}E)=\bigoplus_{k=0}^n\Lambda^k(T^{*}E)$ a complex line bundle over $T^{*}E$?
There seems to be a lot of confusion in your question. Are you sure that is what you mean to ask? As I understand it, the answer is trivially no. A vector bundle that decomposes as a direct sum of vector bundles has dimension equal to the sum of the dimensions of the vector bundles that appear in the sum...
Apr
18
answered Klein Bottle Embedding on $\mathbb{R}^4$.
Apr
18
awarded  Nice Question
Apr
18
comment Klein Bottle Embedding on $\mathbb{R}^4$.
It's not the antipode you are using. When he applies the antipode of $\Bbb R^3$ to a point of the torus of revolution, one of the circle coordinates, the "horizontal" one, if you see the tours as a vertical circle being rotated along the horizontal circle $C=\lbrace(x,y,z)\mid z=0\text{ and }x^2+y^2=1\rbrace$) is shifted by 180° (which corresponds to the $x\mapsto x+\frac12$) while the second circle coordinate is being "complex conjugated" (which corresponds to the $y\mapsto-y$).
Apr
18
comment Klein Bottle Embedding on $\mathbb{R}^4$.
It is. I'm interested, how does Do Carmo define the Klein bottle?