Jun
14
awarded Nice Answer
Jun
10
awarded Enlightened
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 $T^\ast_{(x,\xi)}(T^\ast Q)$.
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
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
May
11
awarded Electorate
May
11
answered A question on existence of $Spin^c$-structure $P\to M$
May
8
comment How to prove this Weitzenbock formula?
Does this not follow from equation (4.14) on the previous page, which is Lemma 6.1.7 of Donaldson-Kronheimer?
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 $p$?
Apr
1
answered Connection on canonical $\operatorname{Spin}^\mathbb{C}$ spinor bundle on symplectic manifold
Mar
26
answered Understanding $r:\mathfrak{g}\rightarrow Vect(X)$ is the transpose of $d\mu:TX\rightarrow \mathfrak{g}^*$
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