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I came across this link on planetmath and a few facts on that link are confusing me.

According to planetmath, any $1$-parameter subgroup in $GL_n(\mathbb{C})$ arises from the exponential map. That is, for $\phi:\mathbb{R}\rightarrow GL_n(\mathbb{C})$, any one parameter subgroup is of the form $$ \phi(t)=e^{tA}, $$ where $A\in T_e GL_n(\mathbb{C}) \cong M_n(\mathbb{C})$.

Why is the domain for $\phi$ the reals, and not the complexes?

The same link says that the one-to-one correspondence between tangent vectors at the identity and one-parameter subgroups is established via the exponential map instead of the matrix exponential.

What do they mean by the matrix exponential?

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    If $t\in\Bbb C$ would it be $1$-parameter? Think geometrically here.2012-06-30
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    Thank you for the comment anon, but I don't think I see it yet. On paper, I am writing down the following: suppose $s=s_1 + i s_2$ and $t=t_1 + it_2$ where $s_i$ and $t_i$ are reals. I am not sure if I am seeing what you are seeing, and what is supposed to go wrong.2012-06-30
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    Also, the matrix exponential sends matrices to matrices. The exponential map sends a parameter to a curve *using* the matrix exponential. | Think about $\Bbb C$ all by itself - it's a *plane*, clearly not $1$-parameter (which is a curve). For example, $t\mapsto e^{it}$ traces out the circle, but allowing $t\in\Bbb C$, it will trace out all of $\Bbb C^\times$.2012-06-30
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    Here's why I am confused. Here's an example of a $1$-parameter subgroup $\phi:\mathbb{C}^*\rightarrow B$ where $\phi(t)=\left[ \begin{array}{cc} t & 0 \\ 0 & t \\ \end{array}\right]$ and $B$ is the set of invertible upper triangular matrices. Here the domain is $\mathbb{C}^*=\mathbb{C}\setminus \{ 0\}$.2012-06-30
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    In fact that's an isomorphism, but it's not a [$1$-parameter subgroup](http://en.wikipedia.org/wiki/One-parameter_group). A good reason we mean *real* parameter by the term "parameter" is geometric. Again, does $\Bbb C^\times$ look $1$-dimensional to you? It's isomorphic to $\Bbb T\times \Bbb R$ via $z\mapsto (\arg z,\log|z|)$, which is certainly $2$-parameter. If you allow either complex or real parameters then things that are $1$-parameter can also be $2$-parameter; it doesn't make much sense to keep such a convention.2012-06-30
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    That's where I am confused. $\mathbb{C}^*$ has 1 complex dimension and 2 real dimension. So under $\phi$ (whether it lives in $B$ or $GL_n(\mathbb{C})$), it still has 1 complex dimension-- like a complex curve, right?2012-06-30
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    Sure, but you don't want to speak of complex dimensions and real dimensions simultaneously with the mere word "dimension."2012-06-30
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    That's true, I'm just still a bit puzzled why the domain lives in $\mathbb{R}$ while the image set lives as a subgroup over $\mathbb{C}$... I'm re-reading your comments and trying to digest them.2012-06-30
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    Thanks for all the comments anon! They are certainly helpful!2012-06-30
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    Dear math-visitor, One can think of $GL_n(\mathbb C)$ as either a real Lie group (in which case one parameter subgroups have a real variable as their parameter) or as a complex Lie group (in which case one parameter subgruops have a complex varialbe as their parameter). In introductory discussion, the real Lie group point of view is more commonly seen, since it doesn't require the extra baggage of complex geometry. (Not that there is a lot of baggage required, but an average grad students, say, is more likely to have seen a discussion of real Lie groups than of complex Lie groups.) Regards,2012-06-30
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    But viewing $GL_n(\mathbb{C})$ as a complex Lie group, do all $1$-parameter subgroups still arise as $e^{tA}$ where $A\in M_n(\mathbb{C})$ and $ t\in \mathbb{C}$ or is there a possibility for different sort of $1$-parameter subgroups to arise in the complex setting?2012-06-30

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