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Graduate student in mathematics.


Apr
9
comment Explicit form of the foliation associated to a differential one-form
Right, I was being a little sloppy since integral curves usually mean a specific paramterization but for the foliation we just care about the image of the curve in the manifold (as per Lee's comment)
Apr
9
comment Explicit form of the foliation associated to a differential one-form
Correct, the leaves are the integral curves of $x^0\partial_0 + x^1\partial_1$. So explicitly the leaves are parameterized by $x^0 = v^0 t, y^0 = v^1 t$, where $v^0,v^1$ are constants.
Apr
9
answered Explicit form of the foliation associated to a differential one-form
Apr
8
comment Weird definition of Kodaira-Spencer map (What's a relative Kähler differential on a manifold?)
Correct, "dga" means differential graded algebra. This is really a mild Lie algebroid, the point is that when you have some $\mathcal O_X$ module with a Lie bracket (in this case the usual Lie bracket of vector fields) and some compatible action on $\mathcal O_X$ (in this case differentiation of holomorphic functions by holomorphic vector fields), the exterior algebra on the dual has a canonical differential.
Apr
8
answered Weird definition of Kodaira-Spencer map (What's a relative Kähler differential on a manifold?)
Apr
8
comment Weird definition of Kodaira-Spencer map (What's a relative Kähler differential on a manifold?)
The definition of $\Theta_\mathcal{V/W}$? That's eq (1.5) on p. 6 of Advances of in Moduli Theory.
Apr
7
comment Weird definition of Kodaira-Spencer map (What's a relative Kähler differential on a manifold?)
Unfortunately I know hardly any scheme theory so I'm not sure what the corresponding differential geometric notion is. Is there a part on "Advances in Moduli Theory" that references this?
Apr
7
comment Weird definition of Kodaira-Spencer map (What's a relative Kähler differential on a manifold?)
$\Theta_\mathcal{V/W}$ is the sheaf of holomorphic sections of the holomorphic vector bundle that is the kernel of the differential of the projection $T\mathcal V \to T\mathcal W$. What relative differential are you asking about?
Apr
6
comment Working out an example of a Chern class
If you're not set on using the definition of Chern class you mention (as Poincare dual to a zero locus), you can try using the Chern-Weil description (which will give you the rational Chern class, which in your specific example gives you the $\mathbb Z$ Chern class since I'm pretty sure flag manifolds have no torsion in their cohomology).
Apr
5
comment Composition of Irreducible Representation and Surjective Homomorphism
This is more easily proved using the definition of irreducibility.
Mar
26
comment $U(1)$ generators of $SU(2)$
But I would look at the Lie algebra level. Any non-zero element in the Lie algebra will generate, via the exponential map, a $U(1)$ subgroup. These groups will be different if the elements of the Lie algebra are not proportional.
Mar
26
comment $U(1)$ generators of $SU(2)$
If $H_1$ and $H_2$ are two subgroups of $U(1)$ then there exists some $A \in SU(2)$ such that $H_1 = A H_2 A^{-1}$.
Mar
26
comment $U(1)$ generators of $SU(2)$
All $U(1)$ subgroups of $SU(2)$ are conjugate in $SU(2)$.
Mar
26
awarded  Informed
Mar
26
comment Is the determinant the “only” group homomorphism from $\mathrm{GL}_n(\mathbb R)$ to $\mathbb R^\times$?
I'm not sure I see how knowing that $SL$ is the commutator of $GL$ tells you that the trace is the unique form that vanishes on commutators. You could probably prove this though using uniqueness of the Killing form on $\mathfrak{sl}_n(\mathbb R)$.
Mar
26
answered Is the determinant the “only” group homomorphism from $\mathrm{GL}_n(\mathbb R)$ to $\mathbb R^\times$?
Mar
26
comment Is the determinant the “only” group homomorphism from $\mathrm{GL}_n(\mathbb R)$ to $\mathbb R^\times$?
a Lie group homomorphism is determined by the corresponding Lie algebra homomorphism. So what are all of the Lie algebra homomorphisms $\mathfrak{gl}_n(\mathbb R) \to \mathbb R$? These are the elements of the dual space of $\mathfrak{gl}_n(\mathbb R)$ that vanish on commutators. Are multiples of the trace the only ones (the trace is the derivative of $\det$)?
Mar
22
comment Prove the Curvature Tensor is a Tensor
@user13223423 if $T$ is a tensor and $D$ is some differential operator, $T = T+D - D$ is a tensor but neither $T+D$ nor $D$ are tensors. Of course, the sum of tensors is again a tensor. With regarding your other question, tensors can be characterized as those linear operators on sections of a vector bundle that are linear over $C^\infty(M$).
Mar
22
revised Prove the Curvature Tensor is a Tensor
added 1 characters in body
Mar
22
comment Prove the Curvature Tensor is a Tensor
Each of those terms is not a tensor on their own. You also had a typo in the definition of $R$, which I edited.