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Consider a complex submanifold $M$ of a complex ambient vector space $X$. Suppose you have a base point $p \in M$ and a $C^1$ arc $\gamma(t)$ passing through $p$ and staying in $M$, with tangent vector at $p$ denoted by $u$ (the tangent vector is taken by considering $X$ as a real vector space, and $\gamma : [0,1] \rightarrow M$). Is there a complex analytic curve $\tilde{\gamma} : \Delta \rightarrow M$ ($\Delta$ being a complex disk) passing through $p$, whose image is in $M$, with tangent vector at $p$ equal to $u$ (as an element of $T_pM$) ? What if the curve $\gamma$ is only differentiable at $p$ ? Intuitively I think the answer is yes, but I can't construct such a curve $\tilde{\gamma}$. Thanks in advance for your help

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Just work in local coordinates, so you're starting with vector $v\in\mathbb C^k$, and consider the complex line spanned by $v$.

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    $\begingroup$ but is it trivial to see that $v$ is indeed in the tangent plane to $M$ at $p$ ? because that's the whole problem $\endgroup$ – Krys May 17 '13 at 13:43
  • $\begingroup$ Yes. If you think of $M$ as a real submanifold of dimension $2k$, the tangent space is the complex tangent space as a real vector space. What makes this stuff confusing is the introduction of the complexified tangent space with double the dimension: $T_zM\otimes_{\mathbb R}\mathbb C = T^{(1,0)}_zM\oplus T^{(0,1)}_zM$. $\endgroup$ – Ted Shifrin May 17 '13 at 15:30

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