Let $X$ be a complex manifold. I am not sure what people mean when they talk about the cotangent bundle $T^*X$ of $X$. I have two interpretations:

  1. At each point $x\in X$, $T_x^*X$ is the complex vector space dual to the complex vector space $T_xX$, i.e. $T_x^*X$ is the space of all complex-linear maps $T_xX\to\Bbb C$.
  2. At each point $x\in X$, $T_x^*X$ is the dual space to the real vector space $T_xX$, i.e. the space of all real-linear maps $T_xX\to\Bbb R$.

Which one is the right interpretation?

Thinking: I was first thinking that there is a natural isomorphism between the two, but it doesn't seem like so. If I try to get a real isomorphism $$(T_xX)_{\Bbb R}^*\to (T_xX)^*_{\Bbb C},$$ where the first space are the real-linear maps $T_xX\to\Bbb R$ and the second one are the complex linear maps $T_xX\to\Bbb C$ (but viewed as a real vector space), then the isomorphism is always basis dependent.

Note that $$\dim_{\Bbb R}(T_xX)^*_{\Bbb R}=\dim_{\Bbb R} T_xX=2\dim_{\Bbb C}T_xX=2\dim_{\Bbb C}(T_xX)^*_{\Bbb C}=\dim_{\Bbb R}(T_xX)^*_{\Bbb C},$$ so the two spaces are indeed isomorphic. Although I cannot find a basis-independent isomorphism.

  • $\begingroup$ People mean the first thing. $\endgroup$ – Qiaochu Yuan Mar 30 '16 at 14:54
  • $\begingroup$ @QiaochuYuan So the cotangent bundle of a complex manifold is not the cotangent bundle of the underlying real smooth manifold? $\endgroup$ – Paul Mar 30 '16 at 14:55
  • $\begingroup$ Hmm. Maybe I confused myself. $\endgroup$ – Qiaochu Yuan Mar 30 '16 at 15:27
  • $\begingroup$ Suppose you have a basis for $V_\mathbb{R}$ or $V_\mathbb{C}$, how do you construct the isomorphism ? $\endgroup$ – user90041 Apr 8 '16 at 10:33

If $V$ is a vector space over $\mathbb{C}$, there is a natural $\mathbb{R}$-isomorphism $\varphi\colon V_{\mathbb{R}}^*\to V_{\mathbb{C}}^*$ defined by $$ \varphi(f)(v) \,=\, f(v) + i\,f(-iv). $$ for $f\in V_{\mathbb{R}}^*$ and $v\in V$, with inverse $\varphi^{-1}\colon V_{\mathbb{C}}^*\to V_{\mathbb{R}}^*$ defined by $$ \varphi^{-1}(g)(v) \,=\, \mathrm{Re}\bigl(g(v)\bigr). $$ Note that if $f\colon V \to \mathbb{R}$ is $\mathbb{R}$-linear, then $\varphi(f)$ is indeed $\mathbb{C}$-linear, since $$ \varphi(f)(iv) \,=\, f(iv) + i f(v) \,=\, if(v) - f(-iv) \,=\, i\bigl(f(v) + if(-iv)\bigr) \,=\, i\,\varphi(f)(v) $$ for any $v\in V$.

  • $\begingroup$ Very nice! Had you encountered this $\phi$ before? $\endgroup$ – Jason DeVito Apr 1 '16 at 4:46
  • 1
    $\begingroup$ @JasonDeVito Not as such, but in general it's possible to take a real-valued harmonic function and add $i$ times its harmonic conjugate to obtain a holomorphic function. This is actually a special case of that, where $f(-iv)$ is the harmonic conjugate of the linear functional $f$. $\endgroup$ – Jim Belk Apr 1 '16 at 12:57
  • $\begingroup$ That just shows my complex analysis is rusty - I should have known that. Thanks! $\endgroup$ – Jason DeVito Apr 1 '16 at 12:58

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