Henry T. Horton
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 Feb23 awarded Enlightened Feb23 awarded Nice Answer Feb13 awarded Yearling Feb8 awarded Nice Answer Dec20 awarded Constituent Dec17 awarded Good Answer Dec10 awarded Caucus Sep30 awarded Explainer Sep13 awarded Good Answer Jun14 awarded Nice Answer Jun10 awarded Enlightened Jun10 awarded Nice Answer May19 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 $V \otimes_{\Bbb R} \Bbb C$. May19 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 $\{dz, d\bar{z}\}$). May19 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 $i$. May17 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 $N_j$. May17 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 $T^\ast_{(x,\xi)}(T^\ast Q)$. May17 revised How many differential forms on the complex plane? edited body May17 answered How many differential forms on the complex plane? May16 answered the tautological 1 form