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I am reading P.M.H. Wilson's Curved Spaces and would be grateful for some clarification.

Firstly, on page 77,

Of particular importance will be the case when $U$ is an open subset of $\mathbb R^n$ and $f:U\to \mathbb R$ is smooth. For each $P\in U$, we have the linear map $df_P:\mathbb R^n\to\mathbb R$ (an element therefore of the dual space to $\mathbb R^n$). These then yield a smooth map $df:U\to$ Hom$(\mathbb R^n, \mathbb R)$, where Hom$(\mathbb R^n, \mathbb R)$ denotes the dual vector space, consisting of homogeneous linear forms on $\mathbb R^n$, where the dual space may also be identified with $\mathbb R^n$.

I cannot understand what it is saying. I have tried looking up dual space but it's all Greek to me. Also why is this statement true? I understand that if $f:U\to \mathbb R$ is smooth, then we can differentiate it, but why is the resulting $df_P$ necessarily linear and why is $df$ homogeneous? (In my understanding, homogeneous means $f(ax)=a^nf(x)$, for some $n$ right?)

Secondly, on the same page,

$f(x+iy)=u(x,y)+iv(x,y)$ is an analytic function of $z=x+iy$ if and only if the Cauchy-Riemann equations are satisfied.

I remember being told once that there are exceptions, in that all analytic $f$'s satisfy CR but satisfying CR is not sufficient to claim that $f$ is analytic. Is my memory faulty? When does it fail? If I am not mistaken, why can we assume this to be true? Are the example that fail very exotic hence ignorable?

Thank you for your time.

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The dual space of a vector space $V$ is the space $V^*$ of bounded linear functions $f:V\to \mathbb F$, where $\mathbb F$ is the base field of $V$. In the case of finite-dimensional vector spaces, the dual space is isomorphic to $V$. A typical example is $V=\mathbb{R}^n$, the elements of which are generally written as column vectors. The elements of the dual space are generally written as row vectors, so that they act as linear functions by matrix multiplication. In this case it is clear that transpose is an isomorphism from $V$ to its dual. –  Alex Becker Feb 21 '12 at 9:23
    
@Pete: If you consider functions which are not continuously twice differentiable on a neighbourhood to be exotic, then there will only be exotic counterexamples to "CR implies analytic"... –  Zhen Lin Feb 21 '12 at 17:21
    
@AlexBecker: Thanks for replying. –  Pete Feb 22 '12 at 11:58
    
@ZhenLin: Thanks for commenting. Hmmm... So they really aren'tall that exotic afterall? –  Pete Feb 22 '12 at 11:58
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