Talk:Surface (geometry)

Article Checklist for "Surface (geometry)"

Workgroup category or categories	Mathematics Workgroup [Categories OK]
Article status	Stub: no more than a few sentences
Underlinked article?	Yes
Basic cleanup done?	Yes
Checklist last edited by	--AlekStos 14:46, 26 March 2007 (CDT)

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Are surfaces necessarily infinite?

Is a surface only something like a plane or sphere, which has no edges, or would one face of a cube count as a surface? If a surface necessarily has no edges, isn't it misleading to say it has length and breadth? --Catherine Woodgold 20:20, 27 April 2007 (CDT)

I think the point here is that surfaces are 2-dimensional. In algebraic geometry, you can consider surfaces defined over arbitrartry fields (even finite ones), but in differfdential geometry you're pretty much limited to R (or C). But even in the case of algebraic surfaces, you usually work over an algebraically closed field and then talk about points definable (or "rational") over a subfield. Greg Woodhouse 21:37, 27 April 2007 (CDT)

I think there are different definitions of surface used in different branches of mathematics, thus the confusion. What the article currently says doesn't seem to me to agree with what you're saying.

I think the article needs to be expanded to cover several different definitions, including at least one mathematically rigourous one (probably provided by someone other than me). It could mention definitions from physics or whatever, at least in order to clarify that that's not what's meant here.

At the moment it looks ambiguous to me. --Catherine Woodgold 10:26, 29 April 2007 (CDT)

That's a tough one. The usual approach is to require that there be mappings (functions) of the form $\scriptstyle \phi :U\subset \mathbb {R} ^{2}\rightarrow S$, sometimes called coorinate charts, such that if $\phi$ and $\psi$ are two charts about the same point, the composite $\scriptstyle \phi \circ \psi ^{-1}:\mathbb {R} ^{2}\rightarrow \mathbb {R} ^{2}$ is a differentiable map with a differentiable inverse. This, in turn, is true if the Jacobian matrix has nonzdero determinant (by the inverse function theorem for functions of several variables). Intuitively, all this means is that the change of local coordinates doesn't "collapse" anything as would, say, $\scriptstyle (x,y)\mapsto (x,0)$. That's a lot of jargon. Feel fre to translate it all into English! Greg Woodhouse 11:49, 29 April 2007 (CDT)

Sorry, I can't translate that into English because I don't understand it! I don't know the meaning of "charts", I don't know what S is -- oh, wait! S must mean the surface, and charts are just functions such as you describe. $U$ is a subset of $\mathbb {R} ^{2}$ (the Euclidean plane) and $\phi$ is a function from U to S. If U is only a subset of $\mathbb {R} ^{2}$, this has me wondering again whether limited surfaces such as one face of a cube would count. I think you've abbreviated a lot -- it's not easy to follow. OK, you mean you can find at least one chart $\phi$ and at least one chart $\psi$, which could both be the same chart (I don't see why one wouldn't usually use the same one), such that if you apply one to get from $\mathbb {R} ^{2}$ to the surface S and then the inverse of the other to get back to $\mathbb {R} ^{2}$, you have a differentiable mapping from $\mathbb {R} ^{2}$ to $\mathbb {R} ^{2}$ with a differentiable inverse. I'm not sure why it has to be required to be differentiable. Just requiring that both be functions (rather than multivalued relations) would seem to prevent planes from collapsing onto lines. I think to get a Jacobian matrix in the first place you have to be dealing with a differentiable function. I think the Jacobian in this case is a matrix of four partial derivatives. Maybe you mean the Jacobian of $\scriptstyle \phi \circ \psi ^{-1}$. Maybe you mean that if it has a nonzero determinant then the inverse of that function exists and is also differentiable. In general, if working with very non-Euclidean geometries I'm not sure whether "determinant" would be replaced with some other norm. I think this definition doesn't allow any surfaces to exist in finite fields. I'm sort-of thinking out loud here. I think I almost understand it. --Catherine Woodgold 13:17, 29 April 2007 (CDT)