Henry T. Horton
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 Jun 14 awarded Nice Answer Jun 10 awarded Enlightened Jun 10 awarded Nice Answer 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 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.