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Let $r=V(0,1)$ be the four-dimensional spinor representation. The tensor product representation splits into its symmetric square and exterior square, with dimensions ten and six. If $v$ is a highest-weight vector of $r$, then $v\otimes v$ is a highest-weight vector of weight $(0,2)$ in the symmetric square. The irreducible representation with that highest weight has dimension ten by solution, exhausting the symmetric square:For the exterior square, add each pair of distinct weights from solution. The resulting multiset of weights isIts maximal weight $(1,0)$ is the highest weight of a five-dimensional irreducible representation. It accounts for the four nonzero weights and one zero weight, leaving a single zero weight. By complete reducibility of semisimple Lie algebra representations, that remaining one-dimensional summand is the trivial representation. Thus ${\Lambda }^{2}r\cong 5\oplus 1$, and the spinor tensor square for so5 isThe dimension check is $16=1+5+10$. The trivial representation lies in the exterior square, reflecting the invariant nondegenerate alternating bilinear form of this pseudoreal representation. Although $r$ itself does not descend to $SO(5)$, its tensor square does: the central element $-1\in \mathrm{Spin}(5)$ acts as $(-I_4)\otimes (-I_4)=I_{16}$.

Use the basis matrices $M_{ab}=E_{ab}-E_{ba}$ from solution. Define the Pauli matrices and the two-by-two identity byThe following five Hermitian four-by-four matrices, written as Kronecker products, realize a Euclidean Clifford algebra:Define the Lie algebra representation on ${\mathbb{C}}^4$ byThese are anti-Hermitian matrices, appropriate to the real compact Lie algebra. For complete explicitness, all ten representation matrices areEach table entry is the image under $\rho$ of the basis matrix immediately to its left. Since the two-by-two factors are specified above, these Kronecker products specify every entry of each four-by-four matrix.

To verify the representation law, the Clifford algebra relations implyApply this identity to $\rho (M_{cd})=\frac{1}{4}[{\gamma }_c,{\gamma }_d]$ and use the commutator product rule. The result isexactly the Lie bracket relation in solution. Thus $\rho$ is a Lie algebra representation.

It remains to identify its highest weight, rather than just its dimension. The commuting Hermitian Cartan subalgebra operators areTheir simultaneous eigenvalues are $(x,y)=(\pm \frac{1}{2},\pm \frac{1}{2})$. Converting with $(a,b)=(x-y,2y)$ gives exactly the four weights of $V(0,1)$ in solution. A state with $(x,y)=(\frac{1}{2},\frac{1}{2})$ has highest weight $(0,1)$, so this Clifford realization of the so5 spin representation is the requested $r$.

Finally, $\rho (M_{12}{)}^2=-I_4/4$, soThis explicitly exhibits the distinction between the spinor representation of $\mathfrak{so}(5)$, or of $\mathrm{Spin}(5)$, and a single-valued group representation of $SO(5)$.