# Tag Info

14

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 ...

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 ...

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

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 $\phi:\mathbb C[x_1,x_2,x_3,x_4]/\mathfrak p\to \mathbb C[s, t]$. ...

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

5

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

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

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 ...

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

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 ...

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

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 ...

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

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 $A-I_4=\begin{bmatrix} 0 & 0 & 0 & -2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ -1 ... 3 The statement with the hypotheses given in Hartshorne is true. For a reference, see COR 13.4 on pg. 290 of Eisenbud's Commutative Algebra. The general idea of proof is this: Consider a maximal chain of prime ideals in$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

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

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

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

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 ...

2

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 $$P=\langle x_1x_3-x_2^2,x_2 x_4-x_3^2,x_1x_4-x_2 ... 2 I'll prove the following result:$$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],$$where K is a field. Let \varphi: K[x_1,x_2,x_3,x_4]\to K[s, t] be the ring homomorphism that maps x_1\mapsto s^3, x_2\mapsto s^2t, x_3\mapsto st^2 and x_4\mapsto t^3. Obviously ... 2 I still don't understand your notation. Neverthless the answer to the question in the title is no. Example: Let R=\mathbb C[x_1,x_2], let I' be the ideal of R[t] generated by x_1^2+t-1. Then R[t]/I'=\mathbb C[x_1, x_2] is flat over \mathbb C[t]=\mathbb C[x_1^2] (it has no torsion). The fiber over 0 is two copies of the affine line and is ... 2 By induction! We claim that \chi(n,\mathfrak{a}+(f_1,\ldots,f_k)) has the same leading coefficient as \chi(n,\mathfrak{a}) for any k\le d and generic (linear) choice of the f_i. For k=1, the statement follows from the result you quoted:$$\begin{align*} \chi(n,\mathfrak{a}+(f))&=\chi(n,\mathfrak{a})-\chi(n-1,\mathfrak{a}) \\&= ...

2

You can again use induction to prove this fact. Base $pd_R M=0$ is obvious. If $pd_R M=n>0$, let $F$ be a finitely generated module s.t. we have an epimorphism $F \to M$, denote kernel of this map $K$, so we have short exact sequence $$0 \to K \to F \to M \to 0$$ but $Tor_1(M, R/x)=\{a \in M : xa=0\}=0$, thus the sequence over $R/xR$ $$0 \to K/xK \to ... 2 If I understand the notation correctly, then (0 :_{M} \mathfrak m^t) denotes the submodule of M consisting of elements killed by \mathfrak m^t. Call this submodule N. Consider the short exact sequence$$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 $M/N = ... 2 The monotonicity of dimension functions is tricky for quite general spaces. For$\operatorname{ind}$we have the easiest situation: for every subspace$A$of a regular space$X$(most texts only consider$\operatorname{ind}$to be defined for regular spaces)$\operatorname{ind}(A) \le \operatorname{ind}(X)$. The dimension functions$\operatorname{Ind}(X)$... 2 The tesseract is the four dimensional analog of the cube. It lives in$\mathbb{R}^4$, four dimensional euclidean space. This four dimensional space has all dimensions equivalent, with none of them being special like time. The space is the set of points$(x,y,z,w)$where the coordinates range over the reals. One of the tesseracts has$16$vertices, with ... 2 If you impose$m$linear and linearly independent constraints on$\mathbb R^n$, then the set has the dimension$n-m$. The vector space of these$m$constraints has the dimension$m$. If only$k, k \le m$(and not more) of these constraints are linearly independent, then the subspace of the constraints has the dimension$k\$ and the orthogonal subspace has ...

Only top voted, non community-wiki answers of a minimum length are eligible