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Oct
11
comment Are Christoffel symbols invariant under reparameterization of the curve?
firstly, term in the Hamilton-Jacobi equations for a geodesic has $\dot \sigma$, not $\sigma$, secondly, the question as it is asked certainly has a negative answer. Surely, Cristoffel symbols only depend on the metric and the chosen local coordinate system, but think of a question to the one you asked: whether $A x(t) = A x(s(t))$ for A some fixed matrix; if your reparametrization is not $s=t$ then you get a different function a priori.
Sep
25
comment Is a bijective morphism of quasi-affine smooth varieties an isomorphism?
@Roland: yes, the statement is really about a bijection on $k$-points of $k$-varieties
Sep
25
comment Is a bijective morphism of quasi-affine smooth varieties an isomorphism?
@Heitor: I remember that there is a proof in Raynaud's book "Anneaux locaux henseliennes", but I expect an exposition in English to be in Stacks project by now (haven't checked it though).
Sep
25
comment Is a bijective morphism of quasi-affine smooth varieties an isomorphism?
@arthur: this equation defines a (non-singular) elliptic curve, which surely cannot be isomorphic to $\mathbb{A}^1$
Sep
25
answered Is a bijective morphism of quasi-affine smooth varieties an isomorphism?
Sep
23
awarded  Disciplined
Sep
21
awarded  Yearling
Aug
19
comment why is there no non-degenerate 2-forms on 4-sphere?
thanks! funny, I remember asking this question because I tried to prove that spheres apart from $S^2$ and $S^6$ do not admit an almost complex structure.
Aug
17
comment Let G be an abelian group, and V be a faithful irreducible representation of G over C
perhaps jackwo meant "preserves". but any such representation would be 1-dimensional anyway
Aug
11
awarded  Autobiographer
Jun
6
comment Kähler differentials, define valuation?
I guess you mean that $R$ is a DVR, so in particular integral. In this case the module of Kahler differentials is trivial, so you can define a valuation on it after choosing some embedding $H^0(X, \Omega_X) \to k(X)$ where $X = \mathrm{Spec} R$, which is tantamount to picking a trivialization. There is no canonical trivialization, but all the valuations obtained this way will be the same since all trivializing sections do not vanish anywhere.
Oct
25
awarded  Nice Question
Sep
28
comment independence of Galois conjugates
Dear Gerry, you are right. I am sorry for the confusion. I have a certain more complicated setting in mind, and I was trying to isolate the question that I am trying to answer, but apparently I was too sloppy formulating it.
Sep
28
revised independence of Galois conjugates
added 26 characters in body
Sep
28
comment independence of Galois conjugates
you are right. I was looking for less trivial examples, though, i.e. something more complicated than Kummer extensions. I guess I should have asked for K to be algebraically closed.
Sep
28
asked independence of Galois conjugates
Sep
26
comment why is there no non-degenerate 2-forms on 4-sphere?
But he claim is for $n\neq 1,3$. Non-degenerate means that restriction of the form to any fibre of the ctangent bundle is non-degenerate as a bilinear form, i.e. induces an isomorphism $T_x M \to T_x^*M$.
Sep
26
revised why is there no non-degenerate 2-forms on 4-sphere?
added 2 characters in body
Sep
26
asked why is there no non-degenerate 2-forms on 4-sphere?
Jul
2
awarded  Curious