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age 28
visits member for 3 years, 9 months
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Math PhD. student and functional programming enthusiast.


Jan
26
answered Differential of the exponential map on the sphere
Jan
25
answered Computation of Laplace-Beltrami operator in a conformally equivalent metric
Jan
25
comment Is Goedel term (in incomleteness theorem) both true and unproveable?
@Suzan: in logic you have two sides: syntactic and semantic. Syntactic is what you can prove formally, while semantic is what is really true (in some model). Ideal theory for a given model should be both sound (you can only prove syntactically what is true semantically) and complete (you can prove syntactically everything that is true semantically). So by adding $\neg G$ you obtain a theory that is not sound anymore (relative to the model for $T$ you were working with previously).
Jan
25
answered Is Goedel term (in incomleteness theorem) both true and unproveable?
Jan
25
comment Is Goedel term (in incomleteness theorem) both true and unproveable?
@Berci: but it is also important to note that extending by $\not \phi$ produces a theory that is not sound since we already know that $\phi$ is true (semantically).
Jan
24
comment $Sp(V)$ acts transitively on $V^*-\{0\}$ where $\Omega$ here is symplectic 2 form
@rschwieb: all of the stuff mentioned in the question is standard and natural. Still, since $\Omega$ is non-degenerate, we can translate all questions about forms to equivalent questions about vectors by $v \mapsto \Omega(v, \cdot)$ and its inverse, so it's a bit puzzling that the author insists on working with $V^*$.
Jan
24
answered normalize subgroup
Jan
24
answered Tangent cone to a subset of $\mathbb{R}^3$
Jan
24
comment Tangent cone to a subset of $\mathbb{R}^3$
Well, one way would be to notice that you can set $z=0$ and work in the $(x,y)$-plane. There the cone will be just two lines and returning back to three dimensions, you obtain the full cone by revolving those two lines around their axis (since $y^2 + z^2 = r^2$ is an equation of a circle in the $(y,z)$-plane).
Jan
24
comment Given any ring $A$, how to obtain $A$ from a ring $B$, in which p is invertible?
Have you heard about localization and are just looking for alternate descriptions? If not, it is possible that looking up information on localization is all you need. In particular, if $R$ is an integral domain and $p \neq 0$ then the ring you are interested in can be obtained as a subring of the field of fractions $K(R)$ of $R$. If $R$ is not an integral domain then one needs to kill the elements $q$ for which $pq = 0$ since otherwise we'd get $q = p^{-1}pq = 0$. I suppose this killing might also be described using standard operations.
Jan
24
answered What is the homology group of the sphere with an annular ring?
Jan
24
answered Minimal Connected Set containing a Closed Connected Set in a Compact Space
Jan
24
comment Minimal Connected Set containing a Closed Connected Set in a Compact Space
@K.Stm.: that doesn't work. Intersection of connected subsets need not be connected. E.g. take $X = S^1$ and $A$ the union two points in $S^1$. Then two arcs containing these points as boundaries are closed and connected but their intersection is just $A$.
Jan
24
comment problem on continuous and differentiable function
Also, as you said, it is not hard to prove this for analytic $f$ (since $f$ behaves like $x^k$ and $f'$ like $x^{k-1}$ near a zero of $f$). I think this generalizes to arbitrary differentiable functions but I can't really come up with the proof. Any ideas?
Jan
24
comment problem on continuous and differentiable function
Looks fine to me. Also, I think it might be possible to show $g(x)$ is actually continuous since otherwise there would need to be an essential discontinuity for $f'$, which means that it would be oscillating a lot and this violates $f'(x) \geq f(x)$ immediately.
Jan
23
comment Show that minimum exists (direct method)
Well, you can generalize this to functional that computes distance along a curve in the plane $\sqrt{1 + (y')^2}dx$ or even further to a functional for a geodesic in a Riemannian manifold or a functional for motion of a point in a potential. The latter is just integral of the Lagrangian $L = T - V$ where $T = v'/2$ is kinetic energy and $V$ is the potential. Your functional corresponds to a free particle in 1D and the solution here is that it stands still.
Jan
22
comment Why does a circle cut a torus into an annulus?
I think you assume that $\phi$ is not contractible in $T^2$ for otherwise, $T^2 \setminus \phi(S^1)$ has two components whereas $T^2 \setminus \iota(S^1)$ has just one and so can't be homeomorphic. But $\psi$ would induce such a homeomorphism.
Jan
22
answered Determine whether the given function is an integrating factor for the DE
Jan
22
comment Adjoint operator on subspace
That's not much better for two reasons. First, $f$ can belong to $H_2 \setminus V_2$, so the expression is again not defined and second, $T$ need not be either injective or surjective, so $T^{-1}$ is not defined in general and even when it is, not for all $f \in V_2$. I think not much can be said about your question unless you add more requirements either on some of the spaces or on the map $T$.
Jan
22
comment What is the last index of a third-order tensor called?
@Giuseppe: I agree with your sentiment of giving names but I stand by what I said. If the question were reformulated so as to use the word matrix instead of tensor (since it moreover doesn't really involve tensors at all -- it's only a question on the terminology of generalized matrices), I will remove the downvote. But I'll leave it here precisely because I am not a fan of the engineering attitude of thinking about the tensor as a matrix (of course, you can treat tensors that way, but then, you can treat almost anything that way, so that's not exactly a great argument..).