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In a vector space $V$ over some field $F$ a hyperplane is the kernel of some linear transformation $T : V \to F$, i.e. the kernel of an element of the dual space (this could be taken as the definition of a hyperplane). Equivalently it is any subspace of codimension $1$, see WolframMathWorld.

But it is well known that for example in $\mathbb R^n$ if we have a linear map $T : \mathbb R^{n-1} \to \mathbb R$, then the set $$ \{ (x_1, \ldots, x_n) \in \mathbb R^n : x_n = T(x_1,\ldots, x_{n-1}) \} $$ gives a hyperplane too, and vice versa every hyperplane in $\mathbb R^n$ could be described this way (quite simple to show, if $H = T^{-1}(0)$ for some $T : \mathbb R^n \to R$, define as $S(x_1, \ldots, x_{n-1}) := x_n$ the unique $x_n$ such that $T(x_1, \ldots, x_{n-1},x_n) = 0$, well-definedness is still left to show, but I guess this is easy, other way, if $H$ is given by such an $S : \mathbb R^{n-1} \to \mathbb R$, define $T(x_1,\ldots, x_{n-1}, x_n) := S(x_1, \ldots, x_{n-1}) - x_n$).

If we introduce coordinates a hyperplane could be described as the solution set of an homogenous linear equation. I guess then the description by a linear map $\mathbb R^{n-1} \to \mathbb R$ is called parameter form, and as the kernel of a map from $\mathbb R^n \to \mathbb R$ it is called coordinate form of the hyperplane

But this distinction is quite unsatifactory for me, because it is in terms of coordinates, and here I am asking if there is some "deeper" definition just in term of abstract vector spaces.

For example, if I define in an arbitrary finite dimensional vector space $V$ a hyperplane as the kernel of an element of the dual space, to what corresponds the parametric form, i.e. the description in terms of a linear map from an $\dim(V)-1$-dimensional space to some space (maybe one-dimensional?).

Is there any more abstract formulation of this correspondence which I sketched in coordinate form.

Remark: By shifting them we get so called affine hyperplanes (the hyperplanes defined here all go through the origin, in contrast to some books where it is not distinguished between hyperplanes and affine hyperplanes).

StefanH
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  • Some spaces in $\mathbb{R}^n$ do not permit us to solve for $x_n$ as functions of the first $n-1$ variables. For example, to keep it simple, consider $\mathbb{R}^2$, there is no way to change $x=1$ into a graph of the form $y=f(x)$. This problem continues. The implicit function theorem tells us in general when we can solve for one of the variables; we need $\partial_k T \neq 0$ in order to solve $T(x_1,\dots , x_n)=0$ for $x_k$. – James S. Cook Jul 18 '14 at 04:25
  • Yes of course, actually I meant solvable for any variable, not necessarily the last. – StefanH Jul 18 '14 at 10:21
  • Of course, in the context of a vector space where we can order the basis however we like then it is possible to solve for the last. In my answer, the last variable is trivially a function of the first $n-1$ variables. That sort of result is also available for abstract manifolds see the last paragraph in the picture in http://math.stackexchange.com/q/345542/36530. It seems wikipedia is lacking a decent article on this... surprising. – James S. Cook Jul 18 '14 at 14:19

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It seems to me that patches or coordinate charts require some choice of basis. However, the real thing which defines a hyperplane is that it has one less dimension than the ambient vector space. However, manifold dimension is not quite restrictive enough as there are plenty of curved spaces of codimension $1$. We should also insist that it is possible to define a vector space structure on a hyperplane. In particular, we could characterize a hyperplane in an $n$-dimensional vector space $V$ as follows:

a hyperplane of $V$ is an embedded submanifold of dimension $n-1$ which is the parallel translate of a subspace of $V$. Alternatively, a hyperplane is a submanifold of codimension one which admits a vector space structure.

In particular, if $H \subseteq V$ is a hyperplane then there exists $v_o \in V$ and $S \leq V$ ($S$ a subspace) for which $H=v_o+S$ where we define $$ v_o+S = \{ v_o+s \ | \ s \in S \}. $$ By linear algebra, there exists $w_1, \dots , w_{n-1}$ a basis for $S$ hence for each $h \in H$ there exist $c_1, \dots , c_{n-1}$ such that $$ h = v_o + c_1w_1+ \cdots +c_{n-1}w_{n-1} $$ Naturally, $H = T(\mathbb{R}^{n-1})$ where $T$ is the affine map defined in view of the choice of basis for $S$ made above: $$ T(c_1, \dots , c_{n-1}) = v_o + c_1w_1+ \cdots +c_{n-1}w_{n-1}. $$ Can we find a parametrization of $H$ which avoids dependence on the basis for $S$? I don't see a way right off, but, any parametrization of $H$ must be related to the one by some element of $GL(S)$ (the invertible linear transformations from $S$ to $S$). These different choices of parametrization are different patches for the linear manifold $H$.

The inverse of $T$ is the coordinate chart on $H$. Let me introduce the coordinate chart $\Phi_{\beta}$ for $S$ defined by: $$ \Phi_{\beta}(c_1w_1+ \cdots +c_{n-1}w_{n-1}) = (c_1, \dots , c_{n-1}) $$ Then if $h \in H$ we define the coordinate chart as follows: $$ T^{-1}(h) = \Phi_{\beta}(h-v_o) $$ This map gives us a 1-1 correspondence between points in $H$ and points in $\mathbb{R}^{n-1}$.

The Cartesian coordinate form of $H$ is not possible to define as there are no Cartesian coordinates for $V$. Why? Because $V$ is an abstract vector space so coordinates are only possible with a choice of basis. If we make a choice, say by extending $\beta$ to $\gamma = \beta \cup \{ w_n \}$ where $w_n \notin S$ then we can find coordinates for $v_o$ with respect to $\gamma$, $$ \Phi_{\gamma}(v_o) = (0, \dots, 0, b) $$ Thus, the equation of $H$ is just $x_n=b$.

James S. Cook
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