9,484 reputation
2042
bio website
location New Jersey
age 28
visits member for 3 years, 11 months
seen 1 hour ago

Graduate student in mathematics.


Apr
12
answered The kernel of a differential one-form
Apr
12
revised How to differentiate this
deleted 181 characters in body; edited tags
Apr
11
comment What is the kernel of a Maurer-Cartan form?
Oh ok. It looks like they're using a much more general definition of Maurer-Cartan form (for example, their M.C. forms are not even on a Lie group necessarily). It seems like the prototypical example of such a form is a flat connection on a principal $G$-bundle. Note that the kernel of a connection defines a horizontal distribution and this distribution is integrable (and therefore gives rise to a foliation on the total space of the principal bundle) if and only if the connection is flat.
Apr
11
comment What is the kernel of a Maurer-Cartan form?
But do you have an example of where this appears in the foliation literature you mention?
Apr
11
comment What is the kernel of a Maurer-Cartan form?
For 1) It is a vector space valued 1-form so it gives a linear map from vector fields to the Lie algebra. Do you have more context for 2)? I don't see it can have a kernel (viewing it as a map as in 1)) since it takes a tangent vector $v$ at a point $g$ to ${L_{g^{-1}}}_* v$ which is non-zero if $v$ is non-zero since left multiplicaiton by $g$ is a diffeomorphism.
Apr
11
answered Isogenies and dimensions
Apr
10
comment Isogenies and dimensions
Or is your question if $g = g'$?
Apr
10
comment Isogenies and dimensions
Do you mean the linear map $\mathbb C^g \to \mathbb C^{g'}$ is surjective?
Apr
10
comment Can a ring isomorphism change the structure of a module?
Yes that should be true. Though the Clifford algebra is a simpler object then the group algebra of $SO(2n)$.
Apr
10
comment Can a ring isomorphism change the structure of a module?
Ok, I guess you can turn that into an example for a ring if you know about Clifford algebras: take the even Clifford algebra on $\mathbb R^{2n}$. Then the even and odd spinors are two non-isomorphic representations.
Apr
10
comment Can a ring isomorphism change the structure of a module?
Note that if $\phi$ is an inner automorphism then you get an isomorphic module, but in general you don't. I can't think of an immediate example for rings but for groups (where modules are representations) an example is that $Spin(2n)$ has two non-isomorphic spin representations but are related by the outer automorphism of $Spin(2n)$.
Apr
10
comment Notation for the Covariant Derivative of a smooth section in the direction of a tangent vector X
If $\alpha$ is a 1-form and $v$ a vector field, then does the notation $\langle \alpha, v\rangle$ make sense to you?
Apr
10
comment Is the space of smooth sections of a vector bundle finitely generated as a $C^\infty$-module?
I'm guessing you want $M$ to be connected (or at least have finitely many connected components)?
Apr
9
comment Complex line bundle at symplectic manifold
A line bundle and connection with curvature $\omega$ exist if and only if $[\omega/2\pi] \in H^2(M;\mathbb Z)$.
Apr
9
comment Explicit form of the foliation associated to a differential one-form
The leaves are the submanifolds of the form $\{(x^0_0 \exp(t), x^1_0 \exp(t), x^2, x^3) \mid t, x^2,x^3 \in \mathbb R\}$ for some constants $x^0_0, x^1_0$.
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?)