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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 $T^\ast_{(x,\xi)}(T^\ast Q)$.
May
17
revised How many differential forms on the complex plane?
edited body
May
17
answered How many differential forms on the complex plane?
May
16
answered the tautological 1 form
May
16
revised Another differential topology lemma
rolled back to a previous revision
May
12
answered Showing that every finitely presented group has a $4$-manifold with it as its fundamental group
May
12
answered Chern classes via connections
Apr
26
answered A complex manifold isn't a sympletic manifold
Apr
18
revised Why is $[\widetilde{v},\widetilde{w}]_p(f)=0$ when $f$ has a critical point at $p$?
added 489 characters in body
Apr
18
answered Why is $[\widetilde{v},\widetilde{w}]_p(f)=0$ when $f$ has a critical point at $p$?
Mar
26
answered Understanding $r:\mathfrak{g}\rightarrow Vect(X)$ is the transpose of $d\mu:TX\rightarrow \mathfrak{g}^*$
Mar
25
answered Universal Equivariant Line Bundles
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
20
awarded  Enlightened
Mar
20
awarded  Nice Answer
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).
Feb
13
awarded  Yearling
Feb
7
revised Find the equation of two straight lines tangent
edited tags
Feb
2
reviewed Reject Ways to put $5$ balls in $3$ boxes if each box must contain at least $1$ ball.