Vishesh
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 Mar28 awarded Yearling Jan12 awarded Revival Jan1 comment If $f$ is an immersion and $g$ is a submersion, then is $g \circ f$ a local diffeomorphism? I had not put this much thought into it. Thanks a lot, especially for the last bit. Jan1 accepted If $f$ is an immersion and $g$ is a submersion, then is $g \circ f$ a local diffeomorphism? Dec8 awarded Inquisitive Nov11 comment Diffeomorphism between the Grassmannian manifolds $\mathbf{Gr}(n,k)$ and $\mathbf{Gr}(n,n-k)$. Should have said this yesterday. Is the map you asked for just obtained by shifting the last $(n-k)$ vectors in $A$ to the first $n-k$ vectors and then listing the remaining $k$ vectors? Nov10 accepted Diffeomorphism between the Grassmannian manifolds $\mathbf{Gr}(n,k)$ and $\mathbf{Gr}(n,n-k)$. Nov10 comment Diffeomorphism between the Grassmannian manifolds $\mathbf{Gr}(n,k)$ and $\mathbf{Gr}(n,n-k)$. Thanks a lot for your effort. That is very comprehensive!!! Nov10 comment Diffeomorphism between the Grassmannian manifolds $\mathbf{Gr}(n,k)$ and $\mathbf{Gr}(n,n-k)$. @Thanks for your response. Nope. I am not able to explicitly reason this out. I do vaguely understand what you are trying to teach me. But I am still lost mostly. Is $\alpha$ a sort of projection map? and did you mean $E =$?? Nov10 comment Diffeomorphism between the Grassmannian manifolds $\mathbf{Gr}(n,k)$ and $\mathbf{Gr}(n,n-k)$. I was working with the Grassmannian as a homogeneous space obtained by the group action of orthogonal group, actually I need to prove that too. Nov10 asked Diffeomorphism between the Grassmannian manifolds $\mathbf{Gr}(n,k)$ and $\mathbf{Gr}(n,n-k)$. Nov4 comment Help with terminology Thanks. I had given up hope on this one. But can you point me to a specific chapter where this is discussed. Nov4 comment Geometric intuition for the Weingarten map It is the tangent space to the surface $M$ at the point $x$, in other words, space of tangent vectors to $M$ at $x$. Oct25 comment $\lim_{n\rightarrow \infty}n^{-\frac{1}{2}\left(1+\frac{1}{n}\right)}\cdot \left(1^1\cdot 2^2…n^n\right)^{\frac{1}{n^2}}$ Maybe you can still use product of limits although your first limit is wrong. Its simple enough. The second limit in the product can be obtained by using the "sandwich" theorem. Just look at how the term $(1.2^2.3^2 \ldots n^n)^{\frac{1}{n^2}}$ is bounded. Oct25 revised Fixed field of two subgroups of $\operatorname{Aut}_{K}{K(x)}$ improved formatting Oct25 suggested approved edit on Fixed field of two subgroups of $\operatorname{Aut}_{K}{K(x)}$ Oct24 accepted Clarification of notion of proper group action. Oct24 comment Clarification of notion of proper group action. Gosh!! I was just suggested the same book by a friend. Thanks a lot. Just one clarification. Why the big fuss about moving compact sets and not any other kind??Does bringing in compactness facilitate anything??Thanks again!! Oct23 asked Clarification of notion of proper group action. Oct22 comment Linear Algebra Subspace question As Matt said, you seem to have confused the vectors with the components $b_1, b_2, b_3$. All you need to do is just go through the terms and definitions governing this problem, you will be done. I suggest you try and do it yourself.