# $$\mathbb{R}^n$$

Hey Brilliant users! I am writing this note concerning a mathematical "concept" I've been thinking about for at least a year. I was hoping to see if anyone has any insight into this. So here it goes: $$\mathbb{R}$$ can be visualized as a line containing all of $$\mathbb{R}$$. Likewise, $$\mathbb{R}^2$$ can be visualized as a plane with all ordered pairs of real numbers and $$\mathbb{R}^3$$ can be visualized as a "space" containing all ordered triplets of real numbers , as I assume most of you know. 1-D volume is typically called length, 2-D volume is typical called area, and 3-D volume just "volume". My mental block seems to come in when I think of $$\mathbb{R}^n$$ with $$n \geq 4$$. I personally cannot visualize this "space" in any way at all (I'm not sure if I'm just not thinking about it correctly), in other words, I can analytically work with elements/subsets of $$\mathbb{R}^n$$ perfectly well, but when it comes to geometric interpretation, my mind draws a blank. The best way I can convey this is to ask you to consider $$\mathbb{R}^3$$. For the sake of ease of visualization, consider this space with the usual orthonormal basis. Now, in order for a basis to exist for say $$\mathbb{R}^4$$, we must find another vector that is normal to all 3 basis vectors for $$\mathbb{R}^3$$, but geometrically, how does one visualize this vector? It, quite literally, seems impossible for me to visualize.

Another problem I have, which stems from the above, is the concept of volume in $$\mathbb{R}^n$$. Again, it comes down to lack of ability to "see" it. I would appreciate any insight anyone may have regarding how best to visualize this.

Note by Ethan Robinett
3 years, 5 months ago

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Most people have difficultly interpreting $$\mathbb{R} ^ n$$ for $$n = 4$$, so don't even worry about the higher dimensions.

It is often less useful for you to try an think of $$\mathbb{R}^n$$ as a concrete object, because you lack of experience with it will end up hampering your interpretation. This echoes your sentiment of "what is the concept of volume?", and it is not immediately obvious that volume of the unit sphere decreases at higher dimensions. There are certainly strange things that happen in higher dimensions, which cannot be explained with just a 2-D or 3-D interpretation.

I spent a significant amount of time thinking about $$\mathbb{R} ^ 4$$, partly out of curiosity, and also partly to work on the Borsuk Conjecture for $$n =4$$. The most useful visual interpretation that I have had of it, is to think of numerous "thin" copies of space that are laid out one after another. Much like how the cube can be thought of as many copies of a "thin" square that are pasted one after another. With this interpretation, you think of the hypercube as numerous copies of a cube stretched out. Some people like to think about this as the dimension of time, and you are flipping through "pages" or a book.

Staff - 3 years, 5 months ago

Can you give a proof of why the volume of a unit hypersphere decreases as the dimensions increase?

- 3 years, 5 months ago

The proper proof would be to calculate the area of the n-D sphere through integration of the cross-sectional area.
For example, the area of the sphere in 2 dimensions is $$S(2) = \int_{-1}^1 \sqrt{ 1 - x^ 2 } \, dx = \pi$$.
The area of the sphere in 3 dimensions is $$S(3) = \int_{-1}^1 S(2) ( \sqrt{ 1 - x^2 } ) ^ 2 \, dx = \frac{4}{3} \pi$$.
Continue doing so, where your area element is $$S(n-1)$$ scaled by the stretching ratio.

As for an idea of what's happening, for larger dimensions, your balls are actually very spikely. Most of the points would have coordinates whose absolute value lie close to $$\frac{1}{ \sqrt{n} }$$, and few that look like $$(1- \epsilon_1, \epsilon_2, \epsilon_3, \ldots )$$. Thus, most points of the ball lie within a cube of side length $$\frac{2}{ \sqrt{n} }$$, which has area $$\frac{ 2^n } { n^ { n/2} }$$. Unfortunately, the denominator (eventually) grows faster (like when $$n > 4$$ ), and that is why the volume of the cube starts to get really small. You can formalize this argument, by considering what is the number of cubes of side length $$\frac{ 2}{ \sqrt{n} }$$ that you need to cover the sphere, and then show that the total volume tends to 0.

Staff - 3 years, 5 months ago

Consider this: We can draw a line on a piece of paper. We can draw a square on a piece of paper. We can draw a cube on a piece of paper. Why stop there? We can draw a tesseract on a piece of paper. And so forth. With a little bit of effort and practice, you can sort of work out the geometrical properties of 4D figures and even higher. When I try to work out some tesseract problems, I actually do it on a piece of paper.

Tesseract

- 3 years, 5 months ago

Like the 5-D on 2-D problem?

It becomes hard to look at it and understand what is going on though.

Staff - 3 years, 5 months ago

Yeah, it can get difficult past 4D, as you said. But in fact, it's become popular now to use computer animations to "revolve" such higher dimension objects to help "see" the structure. It does take a lot of practice, however. But notice that even that technique makes use of a FLAT 2D screen.

As an example, check out "Calabi Yau" manifolds, which figure prominently in String Theory. Calabi Yau manifolds exists in a 3D space, where each component is a complex number, making it a 6D space. Check out the computer animation here.

- 3 years, 5 months ago

Thank you all for your replies! Clearly, at least for me, imagining this kind of thing will take some getting used to, although, back when I learned multivariable calc I thought the same thing about three dimensional vector spaces, which are now second nature.

- 3 years, 5 months ago

Sorry I am commenting here , a doubt of different topic but in a way related to it , Somewhere I read double integration means volume , so n times integration should mean a geometrical structure of which kind?

- 3 years, 5 months ago

Well you really can't say "double integration means volume" because it really depends on what you're integrating: in some cases volume may be obtained by a single integral, 2 integrals and even 3 integrals. So in short, the answer to your question depends on many things.

- 3 years, 4 months ago