= Solution
Name the four vertices $1,2,\bar2,\bar1$ in their displayed order. This is the <crystal of the defining symplectic representation> in rank two, of <highest weight> $\omega_1$ for the <C2 root system>. Its <weights> are respectively $\varepsilon_1,\varepsilon_2,-\varepsilon_2,-\varepsilon_1$, with <simple roots> $\alpha_1=\varepsilon_1-\varepsilon_2$ and $\alpha_2=2\varepsilon_2$. Each lowering arrow subtracts the corresponding simple root. Thus $\omega_1=\varepsilon_1$ and $\omega_2=\varepsilon_1+\varepsilon_2$.
Use the <crystal tensor-product rule> in which $\widetilde f_i$ acts on the first factor if $\varphi_i(a)>\varepsilon_i(b)$, and on the second factor otherwise. The raising operator $\widetilde e_i$ acts on the first factor for $\varphi_i(a)\geq\varepsilon_i(b)$, on the second otherwise. Here $\varepsilon_i(b)$ and $\varphi_i(b)$ count the raising and lowering steps available in color $i$. The data for the four vertices are
$$
\begin{array}{c|cccc}b&1&2&\bar2&\bar1\\\hline\varepsilon_1(b)&0&1&0&1\\\varphi_1(b)&1&0&1&0\\\varepsilon_2(b)&0&0&1&0\\\varphi_2(b)&0&1&0&0\end{array}.
$$
The strictness difference between raising and lowering ensures that they are inverse along each colored edge. Testing the sixteen ordered tensor vertices gives exactly three vertices killed by both raising <Kashiwara operators>:
$$
1\otimes1,\qquad1\otimes2,\qquad1\otimes\bar1.
$$
Their <highest weights> are $2\varepsilon_1=2\omega_1$, $\varepsilon_1+\varepsilon_2=\omega_2$ and zero.
For completeness, successive lowering generates the following connected components of the <tensor product of crystals>:
* From $1\otimes1$: $1\otimes1$, $2\otimes1$, $2\otimes2$, $\bar2\otimes1$, $\bar2\otimes2$, $\bar2\otimes\bar2$, $\bar1\otimes1$, $\bar1\otimes2$, $\bar1\otimes\bar2$, $\bar1\otimes\bar1$.
* From $1\otimes2$: $1\otimes2$, $1\otimes\bar2$, $2\otimes\bar2$, $2\otimes\bar1$, $\bar2\otimes\bar1$. Their edge colors in order are $2,1,1,2$.
* The vertex $1\otimes\bar1$ is isolated.
They have sizes ten, five and one, are disjoint, and exhaust all sixteen vertices. This proves the <tensor square of the defining C2 crystal> decomposition
$$
\boxed{B\otimes B\cong B(2\omega_1)\sqcup B(\omega_2)\sqcup B(0),\qquad\dim=10+5+1.}
$$
Equivalently, the representation decomposes as $L(\omega_1)^{\otimes2}=L(2\omega_1)\oplus L(\omega_2)\oplus L(0)$. The dimensions agree with the <Weyl dimension formula for C2>, $\dim L(a\omega_1+b\omega_2)=(a+1)(b+1)(a+b+2)(a+2b+3)/6$. The ordered highest vertices would change under the opposite tensor convention, but the irreducible decomposition would be the same.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/iii/paper-3-crystal-components.png]
{title=The ten-, five- and one-vertex components of the C2 tensor-square crystal}
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