Tensor algebra ($T(V)$)

The tensor algebra of a vector space $V$ is the graded algebrawith multiplication given by concatenation of tensors.

The truncated tensor algebra discards products of degree greater than $N$. Its multiplication is $(a\otimes b{)}^{(k)}={\sum }_{j=0}^k{a}^{(j)}\otimes b^{(k-j)}$. Every element with scalar part $1$ is invertible by a finite geometric series in its positive-degree part. Truncated exponential and logarithm are finite polynomials.

The nth tensor power is the iterated tensor product of modules $V^{\otimes n}=V\otimes \cdots \otimes V$.

The tensor square is the second tensor power $V^{\otimes 2}=V\otimes V$. Over a field of characteristic different from two it decomposes as the direct sum of the symmetric square and exterior square.

The symmetric algebra is the quotient of $T(V)$ by the ideal generated by $v\otimes w-w\otimes v$. It is graded by symmetric powers:

The nth symmetric power $S^{n}V$ is the quotient of $V^{\otimes n}$ that identifies tensors differing by a permutation of their factors. Equivalently, over characteristic zero it is the subspace of symmetric tensors.

In the symmetric power ${\mathrm{Sym}}^m{\mathbb{C}}^n$ of the defining unitary group representation, diagonal torus weights are indexed by nonnegative integer tuples summing to $m$, with one monomial per weight space. Torus averaging extracts a monomial from any nonzero invariant complex subspace. The complexified Lie algebra acts by $E_{ij}=x_i{\partial }_{x_j}$ on the symmetric algebra. Whenever $a_j>0$, the displayed action moves one unit of degree from $j$ to $i$ with nonzero coefficient. Finite transfers connect every degree-$m$ monomial, so the subspace is the entire representation. The case $m=0$ is the one-dimensional trivial representation. This proof needs neither the Weyl character formula nor the Weyl dimension formula.

The three-dimensional irreducible representation $V_2$ of SU(2) has weights $2,0,-2$. Thus the character of its $n$th symmetric power is ${\sum }_{a+b+c=n}{t}^{2(a-c)}$. Its weight $2\ell$ has multiplicity $\lfloor (n-|\ell |)/2\rfloor +1$ for $|\ell |\le n$. Subtracting adjacent weight multiplicities gives one copy of each $V_{2n},V_{2n-4},\dots$. The invariant subspace is one-dimensional for even $n$ and zero for odd $n$.

For eigenvalues ${\lambda }_1,\dots ,{\lambda }_d$ of a finite-order linear action, its symmetric power characters satisfy ${\sum }_{n\ge 0}{\chi }_{S^{n}V}{t}^n={\prod }_j(1-{\lambda }_{j}t{)}^{-1}$. The denominator is the alternating generating polynomial of its exterior power characters.

The symmetric square is the quotient of $V\otimes V$ by $v\otimes w-w\otimes v$, and records symmetric tensors of rank two.

For $n\ge 2$, contraction with the invariant symmetric bilinear form splits off the trivial line in the symmetric square of the defining special orthogonal group representation. Its null space is the traceless symmetric square. It contains a highest-weight vector of weight $2e_1$, and the Weyl dimension formula gives $(n+1)(2n-1)$, exactly its dimension. Complete reducibility of compact-group representations therefore makes that null space irreducible. The entire symmetric square is reducible because of the invariant line.

For $V={\mathbb{C}}^{2n}$, $n\ge 3$, the symmetric square splits into an invariant line and its irreducible traceless part of highest weight $2{\epsilon }_1$. The exterior square of the defining orthogonal representation is the adjoint module of highest weight ${\epsilon }_1+{\epsilon }_2$. For $n=2$, the exterior square splits instead into highest weights ${\epsilon }_1-{\epsilon }_2$ and ${\epsilon }_1+{\epsilon }_2$. This exception reflects that so4 Lie algebra is semisimple but not simple.

Over every field, including characteristic two, there is a natural isomorphismThe middle summand maps $v\otimes w$ to the mixed product $vw$. Monomial bases prove bijectivity, and the Leibniz rule makes the map equivariant for Lie algebra representations.

For the defining real representation of $SO(d)$, the symmetric square splits as a scalar trace plus symmetric trace-free tensors: ${\mathrm{Sym}}^2({\mathbb{R}}^d)=\mathbb{R}\oplus {\mathrm{Sym}}_0^2({\mathbb{R}}^d)$. The trace projection is $S\mapsto S-(\mathrm{tr}S)I/d$, so the second summand has dimension $d(d+1)/2-1$.

For $d\ge 3$ the trace-free summand is an irreducible real group representation. To see this, diagonalize a nonzero symmetric trace-free matrix in a nonzero invariant subspace. Two diagonal entries differ. Infinitesimal rotation in their plane yields a nonzero symmetric off-diagonal matrix; rotations carry it to every coordinate pair, and a rotation through $\pi /4$ produces a difference of two diagonal entries. These off-diagonal matrices and diagonal differences span all symmetric trace-free matrices, so the invariant subspace is the whole summand. For $d=24$, its dimension is 299, the third-level nontrivial multiplet of the lowest levels of a fully transverse ND bosonic string.