The latest installment of "Peter finds an interesting idea, spends 1 hour worth of whiteboard lecture on representation theory to his algebra-phobic lover and makes a Mathematica toy in 15 minutes" is brought to you by Prismatoy, a cube that can be collapsed into a parallelpiped:

Basically, we wandered into a puzzle store where he picked one of these up and didn't put it down. (We did pay before leaving!)

I like this because, when restricted to any one of the 6 faces, it gives a visualization of the linear transformation

$$\left[\begin{array}{cc}1 & \cos\theta\\ 0 & \sin\theta\end{array}\right]$$

(up to transformations, scaling and unitary operations) with $0<\theta\leq \frac{\tau}{4}$* being the acute angle in the final configuration. You could derive this quickly at home by imagining one face as a unit square, then exploiting some basic trig to find that the transformation maps (0,1) to ($\cos\theta$,$\sin\theta$), (1,1) to ($1+\cos\theta$, $\sin\theta$), and leaves the bottom side of the square unchanged. The above then follows from knowing how the transformation acts on the standard basis vectors for $\mathbb{R}^2$.

Peter noted that the volume of this structure is given by the area of the base times the height, which, in this case, is the determinant of the linear transformation that takes it from cube form to its parallelpiped shape. To demonstrate this, we name the three vectors along the given three sides $\vec{a}$, $\vec{b}$ and $\vec{c}$:

The area of the base is given by $\lvert\vec{a} \times \vec{b}\rvert$---one can see this because

$$\lvert\vec{a}\times\vec{b}\rvert=\vec{a}\vec{b}\sin\theta,$$

which corresponds exactly to the area in the first picture, except the vectors are no longer of unit length. Now recall that $\vec{a}\times\vec{b}$ is a vector perpendicular to both $\vec{a}$ and $\vec{b}$. Taking the dot product $(\vec{a} \times \vec{b})\cdot \vec{c}$ only takes into account the component of $\vec{c}$ that is parallel to $\vec{a}\times\vec{b}$---in other words, perpendicular to both $\vec{a}$ and $\vec{b}$---in other words, the height of the parallelpiped! Hence taking the magnitude of this quantity gives us base times height, which is volume.

But wait, there's more! The quantity $\left\lvert(\vec{a} \times \vec{b})\cdot \vec{c}\right\rvert$ can be written as

$$(\vec{a} \times \vec{b})\cdot \vec{c}=\sum_{i=1}^3\left(\sum_{j=1}^3\sum_{k=1}^3 \epsilon_{ijk}a_jb_k\right)c_i$$

where (in case the reader hasn't seen it before) the Levi-Civita symbol $\epsilon_{ijk}$ essentially acts as the 'opposite' of the Kroenecker $\delta$ function,

*i.e.*

$$\epsilon_{ijk}=\left\{\begin{array}{ll}1 & i=1, j=2, k=3;\ i=3, j=1, k=2;\ i=2, j=3, k=1\\ 0 & i=k=j\\ -1 & \textrm{else}\end{array}\right..$$

Now imagine taking the determinant of the matrix

$$\left[\begin{array}{ccc} \lvert & \lvert & \lvert\\ \vec{c} & \vec{a} & \vec{b}\\ \lvert & \lvert & \lvert\end{array}\right].$$

I won't put the algebra all out here, but calculating the determinant according to the definition and rearranging it will give the previous nested sum. This technique can also be used to prove that

$$(\vec{a} \times \vec{b})\cdot \vec{c}=(\vec{b} \times \vec{c})\cdot \vec{a}=(\vec{c} \times \vec{a})\cdot \vec{b}.$$

Yay! It is now evident that

$$V_{ppiped} = \left\lvert(\vec{a} \times \vec{b})\cdot \vec{c}\right\rvert = \textrm{det}[\vec{c}\ \vec{a}\ \vec{b}].$$

More generally, the determinant of a matrix is a factor indicating what change in volume (or area, or the appropriate dimensional quality) it produces.

*EPILOGUE:*He spent most of Saturday trying to sketch the manifold of possible shape configurations of this object, then trying to determine whether the set of operations on the toy in SL(3) was a group, then made several demonstrations of these and similar phenomena in Mathematica. However, he did not do the dishes. I cut the lecture for brevity.

*YES, TAU.

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