# Can a cube always be fitted into the projection of a cube?

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?

• 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? Apr 10, 2014 at 19:26
• 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.) Apr 17, 2014 at 22:59
• 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) Apr 18, 2014 at 12:07
• 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. Apr 20, 2014 at 7:34
• 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? Apr 23, 2014 at 0:14

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 $$(v\cdot \eta)\eta=(1/\sqrt{m}\sum_{i=1}^n |l_i|)(1/\sqrt{m}(l_1,\ldots,l_n)),$$ and this vector has norm $$\|(v\cdot \eta)\eta\|=\sqrt{\frac{(\sum_{i=1}^n |l_i|)^2}{m}} \geq 1,$$ 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 $$\|(v \cdot \eta)\eta-(v'\cdot\eta)\eta\|=\|2(v \cdot \eta)\eta\|\geq 2.$$ 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.

• 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.
– user21467
Apr 23, 2014 at 12:11
• 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... Apr 23, 2014 at 13:08
• 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. Apr 23, 2014 at 13:18

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 $B_1^{n-1}$?

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

• 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. Jun 12, 2017 at 13:30