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In the book "Anathem" by Neal Stephenson, Part four begins:

Six weeks after I joined the Edharian order, I became hopelessly stuck on a problem that one of Orolo's knee-huggers had set for me as a way of letting me know that I didn't really understand what it meant for two hypersurfaces to be tangent. I went out for a stroll. Without really thinking about it I crossed the frozen river and wandered...

Now, I'll be frank. I know very little about advanced mathematics, I'm currently in Algebra 2 in high school. I've done some reading on my own about some higher dimensional theory, but most of it is beyond me at the moment. Every time I read this book I always wonder what it means for "two hypersurfaces to be tangent".

Can somebody provide and explanation somebody with my level of understanding could comprehend? (Or am I asking the wrong question? And I'm not really sure what I should tag this as, so please feel free to edit the tags.)

kingsfoil
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    Do you understand what it means for, say, a plane and a sphere to be tangent? If so, you basically know what it means for "two hypersurfaces to be tangent" - the only difference is that while a plane and a sphere "live" in 3 dimensions, hypersurfaces "live" in more dimensions, and thus are difficult to visualize. – senshin May 05 '14 at 15:06
  • Are a plane and a sphere tangent in the same way that a line and a circle are tangent? – kingsfoil May 05 '14 at 16:36
  • Yes, it's the same concept of tangency. – senshin May 05 '14 at 16:40
  • A plane is two dimensional, what is a three dimensional plane? Is that a cube? – kingsfoil May 05 '14 at 17:41
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    Well, a line is a 1-dimensional object living in a 2-dimensional plane. A 2-dimensional plane lives in a 3-dimensional space. As this analogy suggests, the 3-dimensional analogue of a plane is just "3-dimensional space" - like the world we live in, only infinite. (Equivalently, if you made a filled cube infinitely large, that would be the same thing.) Unfortunately, to properly visualize tangency of a 3-dimensional object, you have to be able to visualize the 4-dimensional space it lives in, and that's difficult. – senshin May 05 '14 at 17:45

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The word hypersurface is commonly used as a shortcut for a submanifold of codimension $1$ in a manifold which is called the "background", "ambient manifold". "ambient space" etc.

To speak about tangency one needs to have some structure on the spaces in the consideration. A smooth structure on both the ambient space and the subspace is what one usually would choose for that. But this is not the only way of thinking. The hero of the quoted story may equally use the realm of the algebraic geometry that would make the things more complicated in a way. The algebraic geometers will certainly disagree with me, and I leave up to them to give their own answer.

As a differential geometer, I see this question in the following perspective. Locally, we can represent a regular hypersurface $\Sigma$ inside a smooth manifold $M$ as the zero set of a smooth function $s$ on $M$: $$ \Sigma = \{ p \in M | s(p) = 0 \} $$ with the condition that the differential $\mathrm{d}s$ is a nowhere vanishing linear map at all the points of the hypersurface $\Sigma$ (in the domain of function $s$ if it is only defined locally). Such a function $s \colon M \to \mathbb{R}$ is called a (local) defining function of the hypersurface $\Sigma$.

If two hypersurfaces $\Sigma,\,\Sigma'$ are given (locally) by the defining functions $s,\,s'$ and there is a point $p \in M$ (or it belongs to the domains of both defining functions) which belongs to $\Sigma \cap \Sigma'$, there is a natural notion of tangency of $\Sigma,\,\Sigma'$ at point $p$.

We say that $\Sigma,\,\Sigma'$ are tangent at point $p$ if and only if $$ \mathrm{d} s'(p) = k \cdot \mathrm{d} s (p) $$ for some real number $k \neq 0$, the equality needs to hold only at the point $p$.

This definition is not that far from what an algebraic geometer would use, I guess.

Critical remarks, questions and alternative approaches are more than welcome!

Yuri Vyatkin
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