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May
19
comment How many differential forms on the complex plane?
@GiuseppeNegro I don't know if I would agree with that fact. $z$ and $\bar{z}$ are not vectors in $\Bbb C$. $\Bbb C$ has real basis $\{1, i\}$ and a real basis induces a complex basis in the complexification (although here it might look a little confusing because the basis elements $\{1, i\}$ are not the same as the scalars $1$ and $i$...). A general statement would be that if $\{b_1, \dots, b_n\}$ is a real basis for $V$, then $\{b_1 \otimes 1, \dots, b_n \otimes 1\}$ is a complex basis for $V \otimes_{\Bbb R} \Bbb C$.
May
19
comment How many differential forms on the complex plane?
@GiuseppeNegro $V \otimes_{\Bbb R} \Bbb C$ has real dimension $2 \dim(V)$ and complex dimension $\dim(V)$. This matches your original question: $\Bbb C$ has real dimension $2$, so $\Bbb C \otimes_{\Bbb R} \Bbb C$ has complex dimension $2$ (and a complex basis is given by $\{dz, d\bar{z}\}$).
May
19
comment How many differential forms on the complex plane?
@GiuseppeNegro They are different. We can complexify any real vector space $V$ my taking $V \otimes_{\Bbb R} \Bbb C$, and complexification makes no reference to a complex structure on $J$. For example, $J(dz - d\bar{z}) = i(dz + d\bar{z}) = 2\, dx \neq i(dz - d\bar{z}) = -2\, dy$, so we explicitly see that $J$ is not the same as multiplication by $i$.
May
17
comment Showing that every finitely presented group has a $4$-manifold with it as its fundamental group
@user1770201 SvK yields $\pi_1(X) \cong \langle a_1, \dots, a_{|S|} \rangle$ because $\pi_1(S^{\color{red} 2}) = 1$. You can check using SvK that $\pi_1(X \# Y) \cong \pi_1(X) \ast \pi_1(Y)$ when $X$ and $Y$ have dimension $\geq 3$. As for the image of $c$ in $X_j$, it's really homotopic to $b_j$. Note that $b_j = S^1 \times \{\mathrm{pt}\} \subset N_j$, so the image of $c$ is $S^1 \times \{\mathrm{pt}\} \subset S^1 \times S^2 = \partial N_j$, which is $b_j$ homotoped onto the boundary of $N_j$.
May
17
comment the tautological 1 form
$\pi^\ast$ is the pullback by the projection $\pi: T^\ast (T^\ast Q) \to T^\ast Q$. In this case it is changing where $\xi_i \, dx^i$ lives: on the left, $\xi_i \, dx^i \in T^\ast_x Q$, while on the right $\xi_i \, dx^i \in T^\ast_{(x,\xi)}(T^\ast Q)$. This is why $\tau$ is "tautological": it "doesn't do anything" except move $\xi_i \, dx^i$ from $T^\ast_x Q$ to $T^\ast_{(x,\xi)}(T^\ast Q)$.
Mar
24
comment Universal Equivariant Line Bundles
Yes, and the construction is similar to the nonequivariant case. Let $V$ be the direct sum of countably many copies of each irreducible complex representation of $G$, let $BU_G$ be the space of $1$-dimensional subspaces of $V$, and let $EU_G$ be the space of pairs $(\ell, v)$ where $\ell \in BU_G$ and $v \in \ell$. Then $\pi: EU_G \to BU_G$, $\pi(\ell, v) = \ell$ is a universal $G$-equivariant line bundle.
Mar
2
comment Notations in Riemannian Geometry
$i^\ast TN$ denotes the pullback of the bundle $TN$ via the map $i$: en.wikipedia.org/wiki/Pullback_bundle. It has a different meaning than the map $i^\ast : T^\ast N \longrightarrow T^\ast M$ and in particular does not mean the image of $TN$ under the map $i^\ast: T^\ast N \longrightarrow T^\ast M$ (which does not make sense).
Jan
31
comment Involutive Properties of Space-structures on Smooth Manifolds
There's nothing particularly wrong with Turaev's book. You can read the original Reshetikhin-Turaev papers to see their construction of a 3-manifold invariant. The papers are very readable, especially if you have already been going through Turaev's book. Another reference is Kock's book, but it only treats the case of 2D TQFTs. Kock's book is very easy to read, however, and will give you a good idea of what TQFT is about.
Jan
3
comment Homeomorphism between space of characters and compact set
I don't understand what your exact question is. You've already said what the correct homeomorphism is. Are you asking why $\mathrm{ev}_x \leftrightarrow x$ is a homeomorphism?
Jan
2
comment Christoffel symbols proof
You're one step away. The first and third terms in the last step of your work cancel (the metric tensor is symmetric, and the sums implied by the Einstein summation notation are hence the same).
Dec
17
comment About existence of Morse functions
Looking at ASD's answer, I may have misinterpreted the question. Is $f$ supposed to have only one critical point total, and this single critical point is of index $2$? This is the assumption in ASD's answer. If $f$ can have several critical points, but only one of them can be of index $2$, then my comment applies (note that I assumed I could have an index $3$ critical point).
Dec
14
comment About existence of Morse functions
Do you know anything about handlebodies? Index $k$ critical points correspond to attaching $k$-handles. The only real difference between the handlebody structures of $S^1 \times S^2$ and $\Bbb R P^3$ is how the $2$-handle is attached. You can geometrically cancel the $2$-handle of $S^1 \times S^2$ by attaching a $3$-handle, and then put in a single $2$-handle that gives $\Bbb R P^3$. So from this point of view, there should exist such a Morse function.
Nov
19
comment Delta function proof in QM
@Gerg: Wavefunctions are continuous and continuously differentiable when the potential is an actual function (i.e. not infinite anywhere). The delta potential isn't a function but instead a distribution. Note that a delta potential isn't very physical anyway, so the universe is ok. Check this link for a proof that when $V(x)$ is a function $\psi$ is continuously differentiable, and an alternate proof of your question: quantummechanics.ucsd.edu/ph130a/130_notes/…
Nov
2
comment Can there be non trivial self-dual 5-forms on a 10-dimensional compact orientable manifold without boundary?
Hodge decomposition fails miserably for pseudo-Riemannian manifolds. For example on the torus $T^2$ with the Lorentzian metric $dx^2 - dy^2$, the functions $f_k(x, y) = \sin(kx)\sin(ky)$ are harmonic and linearly independent for all $k$. But $H^0(T^2; \Bbb R) \cong \Bbb R$, so clearly harmonic forms can't correspond the cohomology classes. Note that for a Riemannian metric the Laplacian is elliptic, but in the Lorentzian case it is hyperbolic. The full proof of Hodge's theorem relies on the fact that the Laplacian is elliptic for a Riemannian manifold (i.e. the metric is positive definite).
Nov
1
comment Can there be non trivial self-dual 5-forms on a 10-dimensional compact orientable manifold without boundary?
@Muphrid: No, the wedge product is metric-independent. But the Hodge star does depend on the metric. In dimension $4k + 2$ with a positive-definite metric, $\star$ has no real eigenvectors so the argument produces no contradiction. If you consider the $\pm i$-eigenvectors, then you can get norm zero forms, but that's because you're in a complex vector space and not using a sesquilinear form. In dimension $4k+2$ with a Lorentzian metric, there are $\pm 1$-eigenvectors of $\star$ but now the metric is indefinite, so there can be norm zero forms. There is no contradiction in any case.
Oct
22
comment Surgery on trivial knots
See the edit to my answer. Basically any $3$-manifold whose fundamental group cannot be written as a free product of cyclic groups will not be obtained from surgery on any unlink.
Oct
2
comment Why are histograms being of non euclidean space
Read about information geometry to see the connection: en.wikipedia.org/wiki/Information_geometry
Sep
29
comment Functions from $[E_8]$ to $E_8$
Callus's point also applies to the third question, as the $E_8$ manifold doesn't admit a PL structure either.
Sep
27
comment Normal Bundle is Trivial
@PtF: You want to map $v \in T_p S^n$ to $(p, \langle p, v \rangle)$, where $\langle p, v \rangle$ is the dot product of $p$ and $v$? This won't work, because for any $p \in S^n$, considered as a vector in $\Bbb R^{n+1}$, and any $v \in T_p S^n$, also considered as a vector in $\Bbb R^{n+1}$, $\langle p, v \rangle = 0$.
Sep
23
comment Normal Bundle is Trivial
Yes. A specific isomorphism $f: S^n \times \Bbb R \longrightarrow \nu(S^n)$ is given by $f(p, t) = t \frac{\mathbf{x}}{\|\mathbf{x}\|}$.