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Here's what I have in mind: for $k=\Bbb C$ algebraic equations with $k$-coefficients lead to a $k$-variety in $\Bbb A_k^n$, and that variety inherits the structure of $k$-manifold.

However, the above is not necessarily true when $k=\Bbb H$ is the skew-field of quaternions: the variety doesn't have to have $dim_{\Bbb R}$ divisible by $dim_{\Bbb R}\Bbb H$, let alone a quaternionic structure.

Therefore the question: what properties of (ring, skew-field, ...) $k$ cause a $\Bbb A_k$-variety have a natural manifold structure associate with $k$, whatever it is?

EDIT: based on the comment below I'd like to emphasize that I'm interested in the properties of $k$ that make the phrase "$k$-variety is a $k$-manifold" almost work, modulo singularities and such, or work for "nice enough", i.e. regular varieties.

Here's an example of a hunch: it's possible that the breakdown of "$k$-variety is a $k$-manifold" in the case of quaternions is due to the fact that $\Bbb H$ has "too many" automorphisms. Something along these lines: $k$ needs to have this and that property for its structure to be inherited from $\Bbb A_k$ to $T_p(V)$ for most points $p$ on a generic variety $V$.

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  • $\begingroup$ A variety need not be a manifold if it is singular, e.g. $xy=0$ is not a manifold. $\endgroup$
    – Daniel Loughran
    Commented Jan 13, 2014 at 23:49
  • $\begingroup$ @DanielLoughran: OK, modulo singularities, what properties of $k$ make "regular" $k$-varieties being manifolds? I'm more interested in the properties of $k$ that turn regular points of $k$-varieties into open neighborhoods of $k$-manifold, than in the properties of varieties themselves that make them regular. $\endgroup$
    – Michael
    Commented Jan 14, 2014 at 0:06
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    $\begingroup$ What do you mean by a quaternionic variety? Over a commutative field you can argue that any polynomial is (in a suitable sense) locally linear and hence its zero set is (in a suitable sense) locally a hyperplane, but over a noncommutative field where you allow noncommutative polynomials like $x \mapsto axb$ this fails. $\endgroup$
    – Qiaochu Yuan
    Commented Jan 14, 2014 at 0:40
  • $\begingroup$ @QiaochuYuan: there are at least two possible interpretations of quaternionic polynomials: $\Bbb H[x]$ and noncommutative polynomials in $x$ you mentioned. In the former case assigning value to $x$ is not a homomorphism, so varieties won't make sense; however one can still consider zero-set of noncommutative polynomials. The simplest case I had in mind was the zero-set of $x^2+a=0$, which may consist of a 2-sphere. $\endgroup$
    – Michael
    Commented Jan 14, 2014 at 0:49
  • $\begingroup$ What do you mean by a quaternionic manifold? $\endgroup$
    – Tom Goodwillie
    Commented Jan 14, 2014 at 0:50

2 Answers 2

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About quaternionic analysis:

The naive approach to manifolds needs the implicit function theorem. If you try to mimic complex analysis and define a mapping between quaternionic right vector spaces as quaternionically differentiable if the real derivative is right quaternonically linear, then you end up with quaternionic polynomials of order $\le 1$ (where the coefficients act by left multiplication). This was proved by Fueter around 1900, if I remember correctly. So the resulting theory is extremely rigid.

This is the reason for the development of hypercomplex analysis (which generalizes harmonic functions to the quaternionic setting). Also, from another point of view, for the definition of quaternionic structures on real manifolds.

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Presumably, your field $k$ must be a topological field. To the extent that a $k$-manifold is defined in terms of analytic ideas, this would seem to be necessary.

As you have suggested, the best-developed ideas arise when $k=\mathbb{C}$. I would suggest having a look at Serre's paper GAGA. There, he introduces the analytification functor from complex varieties to complex analytic spaces. From reading this, you might gain some insight into those properties of $\mathbb{C}$ that give such a nice functor.

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