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20

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 ...


10

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


9

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


9

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 ...


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

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 ...


7

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 $K[X_1,\ldots X_n]$ ...


7

Standard self-similarity Fractals are often constructed using a recursive procedure and self-similar sets in particular, are always constructed this way. I think that most folks working in fractal geometry would guess that your picture implies a recursive construction like so: Note that the first set is an initial seed. The second set is composed of ...


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 ...


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

What is the rule or condition to be a 2D or 3D picture? This is a subtle question! In a certain sense, the picture is two-dimensional, since it's displayed on a computer screen. :) That's a trivially literal remark, yet not without mathematical content. A closely related question is, "Does there exist a physical object that looks like the picture from ...


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

For one, it gives an invariant which helps us distinguish sets which are otherwise rather difficult to tell apart, and in a way which is reflected in actual, interesting properties of the sets. On the other hand, there are many situations in which the Hasdorff dimension of a set controls analytical properties of solutions of certain differential equations ...


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

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

$\omega^\omega$ can be visualized, in what I think is a fairly nice way in a static 2D image featured on the wikipedia page for ordinal number: Also, if you're willing to allow dynamic visualizations, then Stephen Brooks's transfinite number line goes well past $\epsilon_0$ (to $\Gamma_0$), as well as providing a more linear (if colorful) look at ...


5

I think Alexander Duality is what you are looking for. I gather that you are a non-expert, so I will attempt to describe in fairly informal terms how Alexander duality deals with the questions that you are interested in. Consequently, I'll suppress the inevitable technicalities, since they don't enter into the very geometric situations that you are ...


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 ...


4

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 ...


4

Choose a basis for $W_1$. This basis can be extended to a basis for $V$ (well-known result). Now let $W_2$ be the span of the newly added vectors. Then $W_1 \cap W_2 = \{0\}$ by linear independence and $W_1 + W_2 = V$.


4

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 ...


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 ...


4

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 ...


4

The connectedness of $[0,1]$ implies that the Lebesgue covering dimension is at least $1$, since an open cover such that every point lies in only one member of the cover must consist of clopen sets, hence by connectedness be the trivial cover $\{[0,1]\}$. To show that the dimension is at most $1$, consider an open cover $\mathfrak{U}$ of $[0,1]$. Since ...



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