Aug 28, 2016 · In general, a hyperplane in $\Bbb R^n$ is an $(n-1)$-dimensional subspace of $\Bbb R^n$. So, in the case of $\mathbb R^4$, you may think of a hyperplane as a rotated version of our three-dimensional space $\mathbb R^3$. In $\Bbb R^3$, a hyperplane is a two-dimensional plane, and in $\Bbb R^2$, a hyperplane is a one-dimensional line.

Sep 11, 2017 · "In geometry a hyperplane is a subspace of one dimension less than its ambient space." However, the Greek prefix hyper- means "'over', usually implying excess or exaggeration". So why do we call a hyperplane a hyperplane, while it actually has less dimensions than the original space? Shouldn't it be called hypoplane rather?

Oct 18, 2020 · Another way to look at this geometrically is with the Notion of scalar product. A hyperplane is the zero locus of a linear functional, as is the space of all vectors orthogonal to the function seen as a vector. (A linear functional is a raw vector, you transpose that and you obtain the vector orthogonal to the hyperplane)

Jun 22, 2022 · Note that the hyperplane itself has zero "thickness", so it is not a subspace: its hypervolume is zero. (It is usually included in one of the subspaces.) When hyperplanes are used as in $\eqref{2}$, they are called half-spaces. (A closed half-space, if it includes the hyperplane; an open half-space, if it excludes the

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First, if we are working over affine space, then the hyperplane is cut out by a global function, so the divisor is principal; in particular, this line bundle is trivial. (In fact, any vector bundle over affine space is trivial, though this is a hard theorem, the Quillen-Suslin theorem .)

Apr 1, 2019 · I'm studying about hyperplanes and their equations and had trouble understanding how exactly the hyperplane equation is to be interpreted. I read this answer and this answer and they are great answers no doubt, but I am still having trouble understanding the equation. The definition of a hyperplane given by Boyd is the set

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Mar 26, 2017 · $\begingroup$ yes, but you can interpret this as hyperplane with v=0 and in such hyperplane coefficient vector u is orthogonal to every vector x-a in this hyperplane. $\endgroup$ – mokebe Commented Mar 25, 2017 at 19:16

So the class of a hyperplane corresponds to all global sections of $\mathcal O(1)$, and hence we can identify these two sets (for our purposes). And since $\mathcal O(1)$ is generated by global sections, they determine $\mathcal (X)$.-Here's an example of what is ment by "generated by a hyperplane section".