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If we project the unit cube, i.e. a axis parallel cube with side length 1 centered at the origin, in $\mathbb{R}^n$ onto a $k$-dimensional subspace of $\mathbb{R}^n$ which contains the origin, can we always fit a $k$-dimensional cube of side length 1 into the projection?

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    $\begingroup$ The projection of your unit cube onto a $k$-dimensional subspace can be seen as $n-k$ projections where you decrease by 1 dimension each time. Hence can you answer your question if you project from $\mathbb R^n$ to an $n-1$-dimensional subspace? $\endgroup$
    – Ian Coley
    Apr 10, 2014 at 19:26
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    $\begingroup$ Here's what I would try, though I have no idea whether it will work. Center the unit cube in $\Bbb R^n$ at the origin. Let $v_1,\dots,v_n$ be the projections, onto the $k$-dimensional subspace, of the basis of $\Bbb R^n$ defined by the edges of the cube. Use Gram-Schmidt to convert this spanning set into an orthonormal basis $w_1,\dots,w_k$ of the $k$-dimensional subspace. Then see if the cube $\{ \sum_{j=1}^k \lambda_j w_k \colon $ each $|\lambda_j|\le\frac12\}$ is contained in the projection. (In dimensions $5$ and above, the obvious norm inequality doesn't work for an arbitrary basis.) $\endgroup$ Apr 17, 2014 at 22:59
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    $\begingroup$ Have you proved this for $n=3$? Does the projection of the unit cube on a plane contains a unit square? (I'm just asking. I don't know the answer) $\endgroup$ Apr 18, 2014 at 12:07
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    $\begingroup$ It seems likely to me that this breaks for large $n$ if we project onto the hyperplane perpendicular to a diagonal, but I'm not sure how to prove it. $\endgroup$ Apr 20, 2014 at 7:34
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    $\begingroup$ Where is the unit cube in this problem? Is it a unit cube with a vertex at the origin, or with its center at the origin? $\endgroup$ Apr 23, 2014 at 0:14

3 Answers 3

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This is unfortunately not a "complete" answer, but there will not be enough space for me to do this in the comments, and at least, it is a partial answer...first, let me just lay down the boundaries here: I will provide an outline for a proof that a unit k-dimensional hypercube will always fit into an ORTHOGONAL projection of a n-dimensional hypercube onto a k-dimensional subspace. And I'll just be assuming the standard Euclidian inner product/norm.

So the first thing to notice is that the corners of a hypercube in $R^n$ is represented by a set of vectors $\{(x_1,\ldots,x_n)\in \mathbb{R}^n: x_i=1/2$ or $x_i=-1/2$ for each $1\leq i\leq n\}$, so in other words all possible permutations of $\{1/2,-1/2\}$ in the components of the vector. Now to make the argument simpler to follow from here on, let's rather work with the hypercube of side-length 2 centered around the origin, then we have: each corner is represented by vectors which are permutations of $\{-1,1\}$.

So the proof I have in mind is an induction proof. We have $\mathbb{R}^n$ with a hypercube of side-length 2 centered about the origin. First consider any 1-dimensional subspace of $\mathbb{R}^n$, spanned by a normal vector $\eta$. Now this vector is of the form $1/\sqrt{m}(l_1,\ldots,l_n)$ where $m=\sum_{i=1}^nl_i^2$. Now since the corner vectors of the unit hypercube consists of all possible permutation of $\{-1,1\}$ at least one of the corner vectors (say $v$) is such that $v\cdot \eta=1/\sqrt{m}\sum_{i=1}^n |l_i|$, in particular $v$ will be the corner vector where the sign of each component agrees with the sign of each component in $\eta$. So we have that the orthogonal projection of $v$ on $\eta$ is \begin{equation} (v\cdot \eta)\eta=(1/\sqrt{m}\sum_{i=1}^n |l_i|)(1/\sqrt{m}(l_1,\ldots,l_n)), \end{equation} and this vector has norm \begin{equation} \|(v\cdot \eta)\eta\|=\sqrt{\frac{(\sum_{i=1}^n |l_i|)^2}{m}} \geq 1, \end{equation} because by expansion of the multinomial $(\sum_{i=1}^n |l_i|)^2 \geq \sum_{i=1}^n l_i^2$.

Now if we take the corner vector $v'$ of the unit hypercube to be the vector such that $v'=-v$, then $(v'\cdot \eta)\eta=(-v\cdot \eta)\eta=-(v \cdot \eta)\eta$ and it follows that \begin{equation}\|(v \cdot \eta)\eta-(v'\cdot\eta)\eta\|=\|2(v \cdot \eta)\eta\|\geq 2. \end{equation} This means that the 1-dimensional hypercube of sidelength 2 will fit inside the projection as required.

Now the induction hypothesis is that for every $k$-dimensional subspace of $\mathbb{R}^n$ ($k\leq n-1$) we can fit a sidelength 2 $k$-dimensional hypercube into the orthogonal projection of the sidelength 2 $n$-dimensional hypercube onto this subspace.

Now take any $k+1$ dimensional subspace of $\mathbb{R}^n$. From here on, its just a sketch of the proof: For such a subspace (let's denote it as $W$) we can find an orthonormal basis, and consequently we can write it as a direct sum of orthogonal complements $W=W_1 \oplus W_1^{\perp}$. In particular we can let $W_1$ be any single vector in our chosen orthonormal basis for $W$. By the induction hypothesis we have that the $k$-dimensional hypercube on $W_1^{\perp}$ fits into the projection of the $n$-dimensional cube onto that space, and by the same argument the $1$-dimensional cube fits into the projection of the $n$-dimensional cube onto $W_1$.

SO NOW, the question is, if we take the $(k+1)$-dimensional hypercube constructed by extruding the $k$-dimensional cube in $W_1^{\perp}$ along the normal vector spanning $W_1$ (in both directions by 1 unit), can we then deduce that it will fit into the projection of the $n$-dimensional cube in $W$. This is where my proof is not complete...I think it comes down to choosing the orthonormal basis in a particular way, i.e. that the vector spanning $W_1$ is parallel to cutlines of the projection space $W$ where it intersects with the hypercube in $\mathbb{R}^n$, but unfortunately I have no rigorous way of defining this, or even to know if it is the best choice. From playing around with a plane intersecting a cube in $\mathbb{R}^3$ in mathematica I am quite convinced though that it is always possible to make a choice of basis on the plane so that this will work...

Now just also as a btw, from playing around with an oblique projection of a unit square onto the y-axis with projection matrix $\begin{bmatrix} 0 & 0 \\ a & 1 \end{bmatrix}$ where $a$ is any number, it actually seems as if orthogonal projection is the "worst case scenario"...so I think this is true for oblique projections as well, but I have no idea how to attempt to prove this.

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  • $\begingroup$ About the $k=1$ case: An easy-to-visualize version of the proof for that case is to note that the cube contains a (Euclidean) ball with diameter equal to the side length of the cube, so the projection of the cube contains the projection of the ball, which is a $k$-dimensional ball of the same diameter. (And when $k=1$, the $k$-dimensional ball is also the $k$-dimensional cube.) ... Still thinking about your proposed inductive step. $\endgroup$
    – user21467
    Apr 23, 2014 at 12:11
  • $\begingroup$ hi @StevenTaschuk . I think there is in fact a problem with the induction step...take for example $\mathbb{R}^3$, projecting onto a plane through the origin. If you take any line in this plane, the orthogonal projection of the cube onto the line as a subspace on its own will generally be larger than the section of the projection onto the plane that covers the line (eg a corner vertex projects orthogonally onto the plane, but the line we are considering does not intersect with this point on the plane necesarily). Well anyway, I hope it is at least some contribution...very interesting problem... $\endgroup$ Apr 23, 2014 at 13:08
  • $\begingroup$ Anyway, I guess the induction step could still work, but then one would have to choose specific subspaces, so that the projection remains orthogonal in the higher dimension. $\endgroup$ Apr 23, 2014 at 13:18
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Not an answer, but a few observations from my vague (possibly wrong) recollections of local theory:

  • It certainly has enough volume to contain the smaller cube. If we project $B_{\infty}^n$ onto the hyperplane $(x_1, x_2, ...\dots, x_n)^{\perp}$ with $\sum x_1^2=1$, then $$Vol(P_{H}B_\infty^n)=Vol(B_\infty^{n-1})\sum|x_i|\geq Vol(B_\infty^{n-1})$$

However that is not enough to contain the smaller cube.

  • By duality, we can look at the $B_1^n$. The ball $P_{H}B_\infty^n$ is the dual ball of a section of $B_1^n$ by a hyperplane. So the equivalent question would be: If we section $B_1^n$ by a hyperplane, can that section be fitted inside $B_1^{n-1}$?
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Sorry but this is not an answer but a further question on this matter. Given the very detailed and Excellent response from Mr Hattingh and give the definition;

Let V be a vector space and W be a subspace of V. Then the orthogonal complement of W in V is the set of vectors u such that u is orthogonal to all vectors in W.

For the definition of the Orthogonal Compliment Projection, my further question is;

If we wish to project a unit cube (with face parallel to the 2-D space of projection can we split the projection into the sum of an Orthogonal projection and an Orthogonal Compliment Projection ?

If so which parts of the 3-D cube would be orthogonally projected and which would be Compliment projected.

I hope this is clear.

Paul Hinrichsen

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    $\begingroup$ The answer section is not to be used for further questions Paul. Either please post a new question or if its a small clarifiaction you can use comments below the answer you wish to clarify. $\endgroup$ Jun 12, 2017 at 13:30

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