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Representation theory describes abstract symmetries by linear maps on vector spaces. A group representation assigns an invertible linear map to each group element, respecting multiplication. The payoff is that a symmetry problem becomes a problem about matrices, invariant subspaces, and eigenvalues. Decomposing the action can reveal combinations of variables that a symmetric system cannot mix.

Take the cyclic group $C_3=\{e,g,g^2\}$, with $g^3=e$. On three real coordinates, let $g$ send $(a,b,c)$ to $(c,a,b)$:This permutation representation does more than relabel a list: it acts linearly on every possible assignment of values to the three positions. The vector $(1,1,1)$ stays fixed. The vector subspace $W=\{(a,b,c):a+b+c=0\}$ is also preserved, since a permutation does not change the sum.

Every vector separates uniquely into these two types:For example, $(4,1,1)=2(1,1,1)+(2,-1,-1)$. The first term belongs to a trivial representation; the second is a contrast with zero coordinate sum. This is a direct sum decomposition ${\mathbb{R}}^3=\mathbb{R}(1,1,1)\oplus W$. The symmetry preserves the mean and rotates the contrasts.

The rotation is literal after choosing the orthonormal basis $u=(2,-1,-1)/\sqrt{6}$, $v=(0,1,-1)/\sqrt{2}$ of $W$. In this basis,It is a rotation through $120$ degrees, and three applications return to the starting point. There is no nonzero invariant real line, because this rotation has no real eigenvector. Thus the two-dimensional action is an irreducible representation over the real numbers. Over the complex numbers it instead splits into two eigenlines with eigenvalues $e^{2\pi i/3}$ and $e^{-2\pi i/3}$. The coefficient field is part of an irreducibility claim.

Why would this decomposition matter for a physical calculation? Consider three equal masses with scalar displacements $x_i$ and the symmetric quadratic coupling energyThe equations of motion are $m\ddot{x}=-kLx$, whereis a graph Laplacian and $J$ is the all-ones matrix. The constant vector has eigenvalue zero: translating all coordinates equally costs no coupling energy. On $W$, $J$ vanishes and $L=3I$, so both independent contrast directions have angular frequency $\sqrt{3k/m}$. A three-coordinate coupled problem reduces to a free mean and two identical harmonic oscillators.

For initial displacement $x(0)=(4,1,1)$ and initial velocity zero, the exact solution is $x(t)=2(1,1,1)+\cos (\sqrt{3k/m}t)(2,-1,-1)$. The fixed mean and oscillating contrast can be read directly from the decomposition. More generally, a linear operator commuting with the group action carries its fixed-vector subspace into itself; with an invariant inner product, it preserves the corresponding orthogonal complement too.

For a finite group, averaging over the action constructs an invariant inner product over the real numbers or complex numbers. The resulting invariant complements explain Maschke's theorem: a finite-dimensional group representation decomposes into irreducible representations. Characters compress the action into traces and help identify those pieces. The assumptions matter; in modular representation theory the coefficient field can have characteristic dividing the group order, and such decompositions can fail. Chapter 4 of Pavel Etingof and coauthors' notes develops these mechanisms. ^[ref].

If a representation has one-dimensional endomorphism algebra, it cannot be a direct sum of two nonzero invariant subspaces: the two coordinate projections would be independent endomorphisms. This obstruction does not require either summand to be irreducible, and does not assert that scalar endomorphisms alone imply irreducibility.

A module is semisimple, or completely reducible, when it is a direct sum of simple modules. Equivalently, every submodule has a direct-sum complement.

A representation is semisimple when it is a direct sum of irreducible representations. Equivalently, every invariant subspace has an invariant complement.

Suppose $R$ is a right Noetherian ring and every essential right ideal contains a regular element of a ring. Preimages of essential regular principal right ideals under left multiplication are essential, so they contain regular elements and prove the right Ore condition. In the classical right ring of quotients $Q$, every right ideal is generated by its contraction to $R$, making $Q$ right Noetherian. An essential right ideal of $Q$ contracts to an essential right ideal of $R$, hence contains a regular element and is all of $Q$. A maximal disjoint complement to any right ideal has essential sum with it, so every right ideal is complemented. The right regular module is therefore a finite direct sum of simple modules, making $Q$ semisimple Artinian.

Suppose the $S_i$ are pairwise nonisomorphic simple modules. Any submodule $W$ of their finite direct sum that projects onto every summand is the entire sum. Induct on $k$. Projection onto the first $k-1$ factors is onto by induction. Its kernel is a submodule of $S_k$, hence zero or all of $S_k$. If zero, $W$ is isomorphic to the first $k-1$ factors but also has quotient $S_k$, contradicting their distinct composition factors and the Jordan–Hölder theorem. Therefore the kernel is $S_k$, proving the assertion. Consequently, if $S_i=A{x}_i$, the map $a\mapsto (a{x}_i{)}_i$ is onto and gives $A/({\bigcap }_i{\mathrm{Ann}}_A(x_i))\cong {⨁}_i{S}_i$ as left modules. The annihilators of these vectors need only be left ideals.

For vector subspaces $U\subseteq W\subseteq V$ of a finite-dimensional vector space, such a $W^{\prime }$ always exists. Extend a basis of $U$ to one of $W$, then extend again to one of $V$. Let $W^{\prime }$ be spanned by the original basis of $U$ together with the vectors added in the second extension. Uniqueness of coordinates gives the displayed intersection and sum. Equivalently, choose a direct sum complement $C$ with $V=W\oplus C$ and put $W^{\prime }=U\oplus C$. Necessarily $\dim W^{\prime }=\dim V-\dim W+\dim U$. In $M_2(\mathbb{R})$, the skew-symmetric subspace $U$ inside the traceless subspace $W$ admits $W^{\prime }=\mathrm{span}\{I,\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\}$.

Construct a symplectic basis two vectors at a time. Choose $e\ne 0$; the nondegenerate bilinear form supplies $g$ with $\omega (e,g)\ne 0$, and set $f=g/\omega (e,g)$. For any vector $x$, remove its paired components byThen $\omega (x^{\prime },e)=\omega (x^{\prime },f)=0$. Thus, for $U=\mathrm{span}\{e,f\}$, the symplectic orthogonal complement gives a direct sum $V=U\oplus U^{\omega }$; its restricted bilinear form remains nondegenerate. Repeat there until no vectors remain, then put all $e_i$ before all $f_i$. This proves even dimension and existence together. Only the nonzero pairing is divided out, so the construction also works in characteristic two. The pair-splitting argument is described in Bilinear Forms by Keith Conrad (Keith Conrad, Bilinear Forms, §5, proof of Theorem 5.1 and Theorem 5.4).

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.

Over a field of characteristic different from two, three linear projections summing to the identity annihilate one another in both orders. Each pair sum is the complementary projection to the third. The sum of two linear projections criterion therefore proves zero products. Every vector is the sum of its three images; applying each projection to a zero sum proves that these images form a direct sum. Choosing a basis in each image gives a simultaneous eigenbasis, with triples of eigenvalues $(1,0,0)$, $(0,1,0)$ or $(0,0,1)$. This is algebraic orthogonality, and does not assume that the projections are orthogonal projections for a specified inner product.

For a module $M$ over an integral domain $R$, its torsion submodule isThis is a submodule. If $rm=0$ and $sn=0$ with $r,s\ne 0$, then $rs(m+n)=0$ and $rs\ne 0$; also $r(am)=a(rm)=0$ for every $a\in R$. The integral domain hypothesis supplies both the nonzero product and commutativity. These are the conventions of ^[ref].

A mixed example is the direct sum $M={\mathbb{Z}}^2\oplus \mathbb{Z}/6\mathbb{Z}$. If a nonzero integer kills $(a,b,\overline{c})$, then $a=b=0$. Conversely, $6$ kills every $(0,0,\overline{c})$. ThusThe torsion submodule isolates the finite-order part while the quotient module keeps the independent integer directions. In general $M/T(M)$ is a torsion-free module: if $r(m+T(M))=0$ with $r\ne 0$, some $s\ne 0$ kills $rm$, so $sr\ne 0$ kills $m$.

The hypothesis matters. In the ring $R=\mathbb{Z}/6\mathbb{Z}$ viewed as a module over itself, the elements $\overline{2}$ and $\overline{3}$ are killed by nonzero scalars $\overline{3}$ and $\overline{2}$, but their sum $\overline{5}$ is a unit and is killed by no nonzero scalar. Thus using arbitrary nonzero scalars in a ring with zero divisors need not give a submodule. A common broader convention uses regular elements of a ring as annihilators; the integral domain convention here avoids that ambiguity.

For a finite-dimensional real or complex linear projection $P$, every vector splits as $Px+(I-P)x$ in the image of a linear map and the kernel of a linear map, respectively. Their intersection is zero, so a basis adapted to this direct sum represents $P$ as $\mathrm{diag}(I_r,0)$. Its trace and rank both equal $r$. In particular, a projection's determinant is zero unless it is the identity. Hermiticity is unnecessary for this algebraic identity; a Hermitian matrix projection is additionally orthogonal.

A smooth map $f:M\to N$ is transverse to an embedded submanifold $S\subset N$ whenThis is a sum of vector subspaces, not necessarily a direct sum. Equivalently the composition of $d{f}_p$ with the quotient to $T_{f(p)}N/T_{f(p)}S$ is surjective. In a slice chart for an embedded submanifold of codimension $k$, the $k$ normal coordinate functions composed with $f$ have surjective derivative on their common zero set. If the preimage is empty the condition holds vacuously.

Choosing a positive system of a root system for a complex semisimple Lie algebra divides its root spaces into negative and positive ones. Their respective sums ${\mathfrak{n}}^{-}$ and ${\mathfrak{n}}^{+}$, together with the Cartan subalgebra $\mathfrak{h}$, give this direct sum of vector spaces. Both ${\mathfrak{n}}^{\pm }$ are nilpotent Lie algebras, and $\mathfrak{h}\oplus {\mathfrak{n}}^{+}$ is a Borel subalgebra. The Poincaré-Birkhoff-Witt theorem turns this into an ordered vector-space decomposition of the universal enveloping algebra, useful for building Verma modules.

If idempotents $e_1,\dots ,e_N$ satisfy $e_i{e}_j=0$ for $i>j$, then the left ideals $A{e}_i$ have an internal direct sum. Multiply a relation $\sum z_i=0$ by $e_1$ on the right to obtain $z_1=0$, and continue in order. Pairwise orthogonality in both directions is unnecessary. If the sum has dimension $\dim A$, it is the whole algebra.

The uniform dimension of a module is the supremum of the numbers of nonzero submodules that can occur in an internal direct sum. Finite uniform dimension is equivalent to having no infinite internal direct sum of nonzero submodules. A finite direct sum of uniform modules contained as an essential submodule gives that number as the dimension. The finite uniform decomposition of a Noetherian module therefore gives finite uniform dimension.

A nonzero module is uniform if every pair of nonzero submodules has nonzero intersection. Equivalently, every nonzero submodule is an essential submodule. A right Noetherian domain is uniform as a right module: if $aA\cap bA=0$ for nonzero $a,b$, the right ideals $b^{i}aA$, $i\ge 0$, form an infinite direct sum, contradicting the ascending chain condition.

A weight-space decomposition expresses a Lie algebra representation as a direct sum of simultaneous eigenspaces for a Cartan subalgebra. Each summand is the weight space for its weight. In complex finite-dimensional representations of a semisimple Lie algebra, the simple coroots act as commuting diagonalizable operators, giving this decomposition.