# Tag Info

16

Yours is a very interesting and subtle question, which often generates confusion. First let us give a name to the property you are interested in: a ring $A$ will be said to satisfy (DIM) if for all $\mathfrak p \in \operatorname{Spec}(A)$ we have $$\operatorname{height}(\mathfrak p) +\dim A/\mathfrak p=\dim(A) \quad \quad (\text{DIM})$$ The main ...

14

Quoting from Abbot's book, "Understanding Analysis": There is a sensible agreement that a point has dimension zero, a line segment has dimension one, a square has dimension two, and a cube has dimension three. Without attempting a formal deﬁnition of dimension (of which there are several), we can nevertheless get a sense of how one might be deﬁned by ...

13

If the person is in a Möbius strip, then it seems we are assuming he is $2$-dimensional. Suppose he has with him two identical circles split into sectors of $120^{\circ}$, and each sector is colored a different color. Notice being $2$-dimensional, he can rotate this circle but not reflect it, so the two circles are identical up to a rotation. Now, let him ...

9

Negative dimension is actually much easier to talk about than complex dimension. Super vector spaces are a natural collection of objects that can have negative dimension; given a super vector space $(V_0, V_1)$ we can define its dimension to be $\dim V_0 - \dim V_1$, and this definition has many nice properties; see this blog post, for example. More ...

9

Your ideal is generated by binomials, so one can be smart about it. There is an algebra map $\mathbb C[x_1,x_2,x_3,x_4]\to \mathbb C[s, t]$ such that $x_1\mapsto s^3$, $x_2\mapsto s^2t$, $x_3\mapsto st^2$ and $x_4\mapsto t^3$, and this map maps your ideal $\mathfrak p$ to zero, so it induces $\phi:\mathbb C[x_1,x_2,x_3,x_4]/\mathfrak p\to \mathbb C[s, t]$. ...

8

The most basic surprise, in my opinion, is that the ratio of the volume of the unit sphere to the volume of the cube circumscribing that sphere tends to 0 as the dimension of the space tends to $\infty$. In other words, a high-dimensional sphere takes up almost no space in the cube that circumscribes it. See pp.4--5 in ...

7

This should never be true for a reasonable definition of dimension (for example the dimension of a manifold). A lower-dimensional thing should have measure zero in a higher-dimensional thing, so removing it shouldn't change the dimension of the higher-dimensional thing. The correct version of the "naive equation" is that the Cartesian product of an ...

7

I don't know if this is on Cover's list, but maybe it should be: For $n=2$ and $3$, any tiling of ${\mathbb R}^n$ by unit $n$-cubes has two with a complete facet in common. But it's not true for $n \ge 10$: see http://arxiv.org/pdf/math.MG/9210222.pdf

6

There are many generalizations of the usual notion of dimension, and they are there to capture different properties. Having said that, the intuition behind the dimension is that it describes the number of degrees of freedom you have, e.g. A point on a line has one degree of freedom. A point on a plane has two degrees of freedom. A point of two-dimensional ...

6

suppose $P_0, \ldots, P_n$ are $n+1$ polynomials, of degree less than $d$. Then by multiplying the $P_i$ among themselves up to $k$ times, you can build at least about $k^{n+1}/(n+1)!$ polynomials of the form $\prod P_i^{\alpha_i}$ of degree less than $dk$. But the dimension of the vector space of polynomials of degree less than $dk$ in $K[X_1,\ldots X_n]$ ...

5

Big surprise: our brains evolved in a three-dimensional environment, and so that is what they are best suited for thinking about. It's easy to visualize because we literally see it all the time. Thinking in higher dimensions is harder because we have no (little?) direct experience with them, so there is not a clear prototype for most people to use as a ...

5

Hausdorff outer measure is defined for all sets, and then we use the definition of Caratheodory to restrict it to a subalgebra of "measurable" sets to get the Hausdorff measure. In $\mathbb R^n$, the $n$-dimensional Hausdorff outer measure is the same (up to a constant factor) as $n$-dimensional Lebesgue outer measure, so they have the same measurable sets ...

5

Here, adapted from an example and a problem in Engelking and with lots of blanks filled in, is an example of a zero-dimensional Tikhonov space with a subspace $-$ in fact a closed subspace $-$ of dimension greater than $0$. The first step is to construct a zero-dimensional Tikhonov space $X$ that is not strongly zero-dimensional; this construction is ...

5

There are formal ways to define dimension, indeed, lots of them, depending on the context. As already noted in a comment, your description of why a circle is one-dimensional is not really correct though. The one-dimensionality of a circle is not a function of the fact that a circle itself can be described by a single number (such as radius), but that to ...

4

Concentration of measure phenomena provide great examples of how our intuition based on low-dimensional space is unreliable in high-dimensions. Compare unit balls in the metric spaces $\mathbb R^n$ endowed with, resp. the Euclidean metric $L_2$, versus $L_1$, and $L_{\infty}$. The unit balls of $L_2$ are bounded by "round" spheres and are sandwiched ...

4

The first thing to say is that there are very few restrictions on a two-dimensional section of the Leech lattice. I will get to those. The jpeg below is such a section. The intersections of the green lines are lattice points, and you can see each green fundamental parallelogram. The blue hexagons are the Voronoi cells. I drew in a bunch of red segments of ...

4

You are correct. There are several ways to show that it is impossible, and it basically boils down to the fact that no open subset of ${\bf R}^n$ is homeomorphic to an open subset of ${\bf R}^m$ if $n<m$ (as $U\cap V$ would be in your case). You can assume without loss of generality that the sets in question are connected. One way to show it is to use ...

4

Chapter 8 of Eisenbud has a short history of dimension in algebraic geometry, even giving axioms for a theory of dimension. The historical order seems to be transcendence degree (think meromorphic functions on a Riemann surface), Krull dimension, then Hilbert functions. In particular, Eisenbud mentions that, though one might suspect differently, the most ...

3

One notion of complex dimension that has been used extensively has to do with self-similar sets. A $t$-neighborhood (i.e. points within distance $t$) of such a set may have volume $v(t)$ bounded above and below by constant multiples of $t^d$, where $d$ is the dimension of the boundary and $t$ is small, but such that $t^{-d} v(t)$ is oscillatory and ...

3

The question is too broad to provide specific examples. If you want a reference, I highly recommend Real-Time Rendering. Chapters that would interest you: Chapter 4, Transforms: Covers matrix transformations and operations, including object rotation. It also covers quaternions which are the usual alternatives to matrices. One of the advantages of ...

3

I'll prove the following result: $$K[x_1, x_2, x_3, x_4]/\left< x_1x_3-x_2^2,x_2 x_4-x_3^2,x_1x_4-x_2 x_3\right>\simeq K[s^3,s^2t,st^2,t^3],$$ where $K$ is a field. Let $\varphi: K[x_1,x_2,x_3,x_4]\to K[s, t]$ be the ring homomorphism that maps $x_1\mapsto s^3$, $x_2\mapsto s^2t$, $x_3\mapsto st^2$ and $x_4\mapsto t^3$. Obviously ...

3

Question 1: It is a theorem that $\mathrm{dim}\ A[x]=\mathrm{dim}\ A+1$ for any Noetherian ring $A$, where $\mathrm{dim}$ denotes Krull dimension. Thus $\mathrm{dim}\ \mathbb C[x_1,x_2,x_3,x_4]=4$, as $\mathrm{dim}\ \mathbb C=0$ trivially. The easiest way to compute the dimension of $R$ is to verify that $$P=\langle x_1x_3-x_2^2,x_2 x_4-x_3^2,x_1x_4-x_2 ... 3 The dimension is two. Note that the vectors  u=\left[ \begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \\ \end{array} \right]  and v= \left[ \begin{array}{c} 0 \\ 0 \\ 1 \\ 0 \\ \end{array} \right]  are in the null space of A-I_4=\begin{bmatrix} 0 & 0 & 0 & -2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ -1 ... 3 We expect a normally-distributed random variable to take values close to the mean, and in low dimensions it does. But in high dimensions, it does not. The volume of a thin hyperspherical shell increases so rapidly as its radius increases that even though the variable has greatest probability density near the mean, most of the probability mass is far from ... 3 These definitions are not the same in general, if M is not f.g. Consider the module \mathbb Q_p/\mathbb Z_p over \mathbb Z_p. Its annihilator is 0, so the first definition gives dimension 1. On the other hand, its support is the closed point of Spec \mathbb Z_p, and so the second definition gives dimension 0. If you are reading an article ... 3 Wikipedia has some useful information. In \mathbb{R}^d, d-dimensional Hausdorff measure is equal to Lebesgue measure (up to scaling). So once you have defined it for Borel sets, you can extend it to Lebesgue-measurable sets in the same way: any Lebesgue-measurable set is of the form A = B \cup C where B is Borel and C is Lebesgue measurable with ... 3 The statement with the hypotheses given in Hartshorne is true. For a reference, see COR 13.4 on pg. 290 of Eisenbud's Commutative Algebra. The general idea of proof is this: Consider a maximal chain of prime ideals in A which includes the given prime \mathfrak p, the length of which is dim A (see Thm A, pg. 290 of Eisenbud). It follows that \dim A ... 3 If I understand the notation correctly, then (0 :_{M} \mathfrak m^t) denotes the submodule of M consisting of elements killed by \mathfrak m^t. Call this submodule N. Consider the short exact sequence$$0 \to N \to M \to M/N \to 0. If we write $I = Ann(M)$ and $J = Ann(M/N)$, then we see that $\mathfrak m^t J \subset I \subset J.$ If \$M/N = ...

3

One way to define the dimension of a finite-dimensional vector space naturally extends to the definition of the Euler characteristic of a bounded chain complex of finite-dimensional vector spaces, and these can be negative. Somewhat relatedly, there is a notion of super vector space which also have a notion of dimension which can be negative. These explain, ...

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