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The rank-nullity theorem says that for a linear map $T:V\to Z$ with finite-dimensional domain,

The direct sum $U\oplus W$ consists of pairs $(u,w)$ with componentwise vector-space operations. The canonical inclusions areFor subspaces $U,W\le V$, defineThis is surjective, andThe First isomorphism theorem for vector spaces therefore gives

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

Maschke's theorem says that every finite-dimensional representation of a finite group $G$ over a field of characteristic zero is completely reducible. More generally, it is enough that the field characteristic does not divide $|G|$.

Schur lemma states the following.

Indeed, $\ker T\subseteq V$ and $\mathrm{im}T\subseteq W$ are invariant subspaces. If $T\ne 0$, irreducibility forcesso $T$ is an isomorphism. For an endomorphism of a finite-dimensional complex representation, choose an eigenvalue $\lambda$ of $T$. The intertwiner $T-\lambda I$ has nonzero kernel, so the first part forcesThis proves the Proof of Schur lemma.

Now let $\rho :G\to \mathrm{GL}(V)$ be faithful and irreducible over $\mathbb{C}$. For every $z\in Z(G)$, the operator $\rho (z)$ commutes with every $\rho (g)$. Schur's lemma therefore givesfor some ${\lambda }_z\in {\mathbb{C}}^{\times }$. Faithfulness makes $z\mapsto {\lambda }_z$ an embeddingSince every finite subgroup of the multiplicative complex numbers is cyclic,This is the cyclic-center obstruction to a faithful irreducible representation.

Let $P_{\bullet }\to \mathbb{Z}$ be a projective resolution of the trivial $\mathbb{Z}G$-module. The projective-resolution definition of group cohomology isThis is independent, up to a natural isomorphism, of the chosen projective resolution.

The degreewise natural isomorphismscommute with the coboundary maps. Taking cohomology proves that group cohomology commutes with finite direct sums:

Now restrict $P_{\bullet }$ from $G$ to a subgroup $K$. The group ring $\mathbb{Z}G$ is free as a $\mathbb{Z}K$-module, so restriction carries free modules to free modules and projective modules to projective modules. Thus the restricted complex is a projective resolution of the trivial $\mathbb{Z}K$-module. For the coinduced module $Y={\mathrm{Hom}}_{\mathbb{Z}K}(\mathbb{Z}G,X)$, the Hom functor adjunction for a coinduced module gives an isomorphism of cochain complexesExplicitly, a map $F$ is sent to $p\mapsto F(p)(1)$; the inverse sends a $\mathbb{Z}K$-linear map $a$ to $p\mapsto (r\mapsto a(rp))$. Taking cohomology proves Shapiro's lemma:

For the conjugation module of a group ring $N=\mathbb{Z}{G}_{\mathrm{conj}}$, the basis $G$ is the disjoint union of its conjugacy classes. Hence $N$ is the direct sum of the integral permutation modules on those classes. The class of a representative $g_i$ is the transitive $G$-set $G/C_G(g_i)$, where $C_G(g_i)$ is its centralizer. Since $G$ is finite, this permutation module is both induced and coinduced from the trivial $C_G(g_i)$-module $\mathbb{Z}$. Applying group cohomology commutes with finite direct sums and Shapiro's lemma yields the group cohomology of a conjugation module:

The symmetric group $S_3$ has three conjugacy classes, represented by the identity, a transposition, and a three-cycle. Their centralizers are respectivelyFor any finite group $L$ acting trivially on $\mathbb{Z}$,because a group homomorphism sends an element of finite order to an element of finite order, while the additive group of the integers contains no nonzero torsion elements. ThereforeThe periodic resolution of a finite cyclic group alternates the maps $t-1$ and $1+t+\cdots +t^{m-1}$. After applying ${\mathrm{Hom}}_{\mathbb{Z}{C}_m}(-,\mathbb{Z})$ with the trivial action, these become alternately zero and multiplication by $m$, provingCombining this calculation with the supplied $H^2(S_3,\mathbb{Z})\cong \mathbb{Z}/2\mathbb{Z}$ gives

The defining two-dimensional group representation is reducible, because the line $\mathbb{R}{e}_1$ is an invariant subspace:For $b\ne 0$, the complementary line $\mathbb{R}{e}_2$ is not invariant, so this representation need not split as a direct sum of one-dimensional representations.

If $T^2=I$, every $v\in V$ has the decompositionwhose first term belongs to $E_1$ and second to $E_{-1}$. Their intersection is zero by ii, so $V=E_1\oplus E_{-1}$ is a direct sum. Choosing bases of the two eigenspaces gives a diagonal matrix with entries $1$ and $-1$.

Since $T^2\ne T$ while $T^2=I$, we have $T\ne I$. Thus $\dim E_{-1}$ is $1$, $2$, or $3$. Reordering the basis gives exactlyFor $-I_3$, $T$ is point inversion through the origin. For $\mathrm{diag}(1,1,-1)$, it fixes the plane $E_1$ pointwise and reverses the complementary line $E_{-1}$: if $v=p+\ell$ in this direct sum, then $Tv=p-\ell$. For $\mathrm{diag}(1,-1,-1)$, it fixes the line $E_1$ pointwise and reverses the complementary plane $E_{-1}$.

These are instances of oblique symmetry of complementary subspaces, allowing oblique eigenspaces. They are a Euclidean reflection in a hyperplane and a rotation in three dimensions through $\pi$ about a line, respectively, when the two eigenspaces are perpendicular. The hypotheses do not require an orthogonal transformation or an orthonormal basis, so that additional metric property cannot be inferred just from the displayed diagonal forms.

If $A\mathbf{v}=\lambda \mathbf{v}$ for a nonzero eigenvector, then $A^2\mathbf{v}={\lambda }^2\mathbf{v}=A\mathbf{v}=\lambda \mathbf{v}$. Thus $\lambda (\lambda -1)=0$, soBoth values need not occur: the zero matrix has only $0$, and the identity matrix only $1$. More precisely, an idempotent linear map whose rank of a linear map is $r$ has $r$ copies of $1$ and $n-r$ copies of $0$. Indeed, its image of a linear map and kernel of a linear map form a direct sum, giving an eigenbasis without assuming an orthogonal projection.

If $Av=\lambda v$ for a nonzero eigenvector, then $A^{2}v={\lambda }^{2}v$. Since $A^2=A$, $\lambda (\lambda -1)=0$: every eigenvalue is zero or one. Let $m$ be the number of ones, counted with algebraic multiplicity. The trace is $m$ and the determinant is the product of these eigenvalues. ThereforeEvery output $y=Ax$ is fixed by a second application:The image of a linear map is thus the eigenspace for the eigenvalue one. To establish its dimension without assuming diagonalizability, decompose every input asThe first term lies in the image of a linear map $A$ and the second in its kernel. Their intersection is zero, because a vector in the image is fixed by $A$ whereas a vector in the kernel is killed by $A$. Thus the space is their direct sum, and an adapted basis puts $A$ into the form $\mathrm{diag}(I_r,0)$, where $r=\mathrm{rank}A$. Comparing traces gives $r=m$:This is an idempotent linear map, and trace of an idempotent equals its rank. If $m=0$, no nonzero output exists. The dimension in the question refers to the image subspace, including its zero vector.

A group representation on a vector space $W$ is a group homomorphism $\rho :G\to GL(W)$: $\rho (gh)=\rho (g)\rho (h)$ and $\rho (e)=I$. An invariant subspace $U$ obeys $\rho (g)U\subseteq U$ for every $g$. A nonzero representation is irreducible if its only invariant subspaces are $0$ and $W$. It is a faithful representation if it is injective, equivalently if its group kernel is $\{e\}$. Its character is ${\chi }_{\rho }(g)=\mathrm{tr}\rho (g)$. For inequivalent complex irreducible representations of a finite group, character orthogonality is

Use right-to-left composition of permutations and assign the matrix labels consistently byThe resulting Cayley table, with row multiplying column, isConjugation relabels the entries of a cycle, preserving its length. Thus the conjugacy classes are

Define the permutation representation by $P_{\sigma }{e}_j=e_{\sigma (j)}$, or $(P_{\sigma }{)}_{ij}={\delta }_{i,\sigma (j)}$. Applying it to a basis vector proves $P_{\sigma }{P}_{\tau }=P_{\sigma \tau }$. The six assignments above produce exactly the listed matrices. They are distinct, and $P_{\sigma }=I$ forces $\sigma (j)=j$ for every $j$, so the representation is faithful.

Every permutation fixes $(1,1,1)$, hence fixes every vector $(c,c,c)$. Therefore $V_0$ is a one-dimensional invariant subspace carrying the trivial representation. Its orthogonal complement issince the inner product with $(c,c,c)$ is $c(v_1+v_2+v_3)$. Permuting coordinates preserves their sum, so this plane is invariant too. Every real vector has the unique decompositionThus the permutation representation is the direct sum of group representations on the line and the plane.

To find the eigenvectors of $A$, note that $A(v_1,v_2,v_3)=(v_3,v_1,v_2)$. With $\zeta =e^{2\pi i/3}$, its three eigenvalues and eigenspaces areThe only real eigenvalue is one, whose eigenline is $V_0$, so $A$ has no nonzero real eigenvector in $V_0^{\perp }$. A proper nonzero real invariant subspace of that two-dimensional plane would be a line, giving such an eigenvector. Hence the plane representation is irreducible over $\mathbb{R}$.

For the later complex dimension-count theorem one should also establish irreducibility over $\mathbb{C}$; real irreducibility alone would not imply it in general. In the complexified plane, the only candidate $A$-invariant lines are its $\zeta$ and ${\zeta }^2$ eigenlines. But $C$ exchanges them:Neither is invariant under the whole group, proving complex irreducibility as well. The required decomposition is thereforewhere $W$ is the standard representation of the symmetric group for three letters. Its character is obtained by subtracting one from the number of fixed coordinates: on the three classes it is $(2,-1,0)$.

Finally, the sign representation gives a second one-dimensional irreducible, with value $+1$ on $I,A,B$ and $-1$ on the transpositions. These are exactly the two one-dimensional possibilities: if $r=A,s=C$, a one-dimensional representation sends them to commuting nonzero scalars $v,u$. The relations $r^3=s^2=e$ and $srs=r^{-1}$ imply $v^3=u^2=1$ and $v=v^{-1}$. Hence $v=1$ and $u=\pm 1$, the trivial and sign choices. Together with the two-dimensional $W$, these giveThe sum of squares of irreducible representation dimensions leaves room for no other irreducible. Thus there are exactly two one-dimensional and one two-dimensional complex irreducible representations.

We first prove that any two nonzero principal left ideals $Ra$ and $Rb$ intersect nontrivially. Suppose instead that $Ra\cap Rb=0$. The nonzero left idealsthen form a direct sum. Indeed, take a finite relationIt impliesAs $R$ is a noncommutative domain and $a\ne 0$, this gives $u_0=0$. Right cancellation of the nonzero element $b$ reduces the remaining relation to one of the same form with one fewer term. Repetition gives $u_1=\cdots =u_m=0$, proving the asserted direct sum.

Consequently its finite partial sums give a strictly ascending chain of left ideals, contradicting the ascending chain condition in a left Noetherian ring. Thus $Ra\cap Rb\ne 0$ whenever $a,b\ne 0$. This is the left Noetherian domain Ore lemma.

Now take $s\in S=R\setminus \{0\}$ and $r\in R$. If $r=0$, choose $s^{\prime }=1$ and $r^{\prime }=0$. Otherwise chooseSince $z\ne 0$, the coefficient $s^{\prime }$ is nonzero and hence belongs to $S$. This proves the left Ore set condition in all cases.

For the submodule $N\subseteq M$, define its subspace filtration and the quotient filtration on the quotient module $Q=M/N$ byThey are compatible filtrations of a module, since multiplication by $F_{i}R$ sends $N_j$ into $N_{i+j}$ and $Q_j$ into $Q_{i+j}$.

In each degree, inclusion and the quotient map induceThe first map is injective: if $n\in N_j$ belongs to $M_{j-1}$, then $n\in N\cap M_{j-1}=N_{j-1}$. The second map is surjective by the definition of $Q_j$.

For exactness in the middle, a class $m+M_{j-1}$ lies in $\ker {\pi }_j$ precisely when $m\in M_j$ and $m+N\in Q_{j-1}$. This means $m=m^{\prime }+n$ for $m^{\prime }\in M_{j-1}$ and $n\in N$. Since $m,m^{\prime }\in M_j$, one also has $n\in N_j$, andConversely, every class from $N_j$ maps to zero in $Q_j/Q_{j-1}$. Taking the direct sum over all degrees gives the short exact sequence of graded modulesThe maps respect the $\mathrm{gr}R$-action because they come from module homomorphisms compatible with the filtrations of a module. Consequentlyas $\mathrm{gr}R$-modules. This proves exactness of associated graded modules for induced filtrations.

Furthermore $A=\mathrm{gr}R$ is a Noetherian ring, and $\mathrm{gr}M$ is a finitely generated module. Its submodule $\mathrm{gr}N$ and its quotient module $\mathrm{gr}(M/N)$ are therefore finitely generated modules. Hence the induced subspace filtration and quotient filtration are both good filtrations for the order filtration.

Let $V\ne 0$ be finite-dimensional. Choose an $H$-eigenvector and apply $e$ repeatedly until the last nonzero vector $v$ in that chain. Such a last vector exists because $e$ raises the $H$-eigenvalue by two and $H$ has only finitely many eigenvalues. Then $ev=0$ and $Hv=lv$ for some $l$.

The nonzero $f$-chain cannot continue indefinitely, since its weights are distinct. Part (a), with its zero endpoint included, gives $l\in {\mathbb{Z}}_{\ge 0}$ and the basis $v,fv,\dots ,f^{l}v$. The raising formula in part (a) shows its span is a submodule isomorphic to $A_l$.

To prove $A_l$ irreducible, a nonzero submodule stable under diagonal $H$ contains a weight vector: polynomial interpolation in the distinct eigenvalues projects onto individual representation weight lines. Repeated raising reaches $v_0$, and repeated lowering then reaches every $v_k$. Thus the submodule is all of $A_l$. Distinct $l$ give different dimensions $l+1$, so the modules are nonisomorphic. Every finite-dimensional irreducible module contains one of these nonzero simple submodules and hence equals it. ThereforeThe zero module has no nonzero simple submodule and is dealt with as the empty direct sum in part (d).

Use the standard defining properties of Category O: finite generation over $U(\mathfrak{g})$, a direct sum of finite-dimensional weight spaces, and locally finite positive root action. In particular the weights lie in a finite union of downward root cones. Submodules and quotients retain the representation weight bounds and finite weight spaces; finite generation of submodules follows from the Poincaré-Birkhoff-Witt theorem and the noetherian property of $U(\mathfrak{g})$.

Choose finitely many weight vector generators. The sum $T$ of their full weight spaces is finite-dimensional and preserved by the centre. The kernel of its central action has finite codimension in $Z(U(\mathfrak{g}))$ and kills the generators, hence, by centrality, all of $M$. Thus only finitely many central characters ${\theta }_1,\dots ,{\theta }_r$ can occur in any subquotient: they are the characters of a fixed finite-dimensional commutative quotient algebra of the centre.

By the Harish-Chandra isomorphism each ${\theta }_i$ determines one finite Weyl orbit of shifted weights. Let $S$ be the finite union of these orbits after subtracting $\rho$. Every nonzero subquotient has a nonzero highest-weight vector, using the bounded-above weights. Within its finite-dimensional highest weight space, the commuting central operators have a common eigenvector. Part (e) and the central-character restriction show that its highest weight belongs to $S$.

Define for any subquotient $N$It is finite, additive in short exact sequences, and positive whenever $N\ne 0$. Hence each strict step of a submodule chain consumes at least one unit of $d(M)$. Refining a nonsimple factor into a proper submodule and quotient cannot continue more than $d(M)$ times. It ends with simple factors, provingThe argument proves the assertion for every object of $\mathcal{O}$, rather than just for a cyclic highest weight object.

Use the intended finite-dimensional irreducible highest weight module $A^{\varphi }$, with dominant integral highest weight $\varphi$. Its highest weight multiplicity is one. An arbitrary direct sum having that same largest representation weight would not in general satisfy this formula. Set $f={\pi }_{A^{\varphi }}$ and use the corrected denominator $D$ of part (a).

Restriction to each simple root ${\mathfrak{sl}}_2$ decomposes into symmetric finite representation weight strings, so $f$ is Weyl-invariant. Consequently $Df$ is Weyl-antisymmetric. We prove that no lower Weyl alternants occur, rather than assuming the desired character formula.

First establish the denominator identity itself. The finite sum $D$ has support in $\rho -Q_{+}$ and coefficient one at $\rho$. Antisymmetry kills all singular representation weight coefficients, since a reflection fixing the representation weight reverses their sign. Every remaining Weyl orbit has a unique regular dominant representative $\eta$. If it occurs in $D$, then $\eta \le \rho$. Since $\eta$ is integral and regular dominant, $\eta -\rho$ is dominant integral; it also belongs to $-Q_{+}$. A dominant vector $v=-\sum n_i{\alpha }_i$ satisfies $(v,v)=-\sum n_i(v,{\alpha }_i)\le 0$, so it must be zero. Thus the only possible orbit is $W\rho$. The leading coefficient givesFor the formal Laplacian $\mathcal{D}(e^{\lambda })=\Vert \lambda {\Vert }^2{e}^{\lambda }$, this implies $\mathcal{D}D=\Vert \rho {\Vert }^{2}D$.

Now choose root vectors dual under an invariant form, so $[E_{\alpha },E_{-\alpha }]=H_{\alpha }$ with $H_{\alpha }$ corresponding to the vector $\alpha$, and take the quadratic Casimirfor an orthonormal Cartan basis. On a highest vector its scalar is $c_{\varphi }=\Vert \varphi +\rho {\Vert }^2-\Vert \rho {\Vert }^2$, hence it has this scalar throughout the cyclic highest weight module.

Here is the computation of the weighted trace of its radial identity. For a positive root, let $X$ be the formal weighted trace of $E_{-\alpha }{E}_{\alpha }$ and $Y$ that of $E_{\alpha }{E}_{-\alpha }$. Cyclicity of trace and representation weight shifting give $Y=e^{\alpha }X$, while their difference is the weighted trace of $H_{\alpha }$, namely $(\alpha ,\nabla f)$. ThusTaking the trace of $C$ yieldsThe product expression for $D$ givesUsing $2\rho ={\sum }_{\alpha >0}\alpha$ and the product rule for the formal Laplacian therefore givesHence every representation weight in the support of $Df$ has norm $\Vert \varphi +\rho \Vert$. Its support is also contained in $\varphi +\rho -Q_{+}$, and it is antisymmetric. Any occurring regular dominant weight $\eta$ thus satisfies $\eta \le \varphi +\rho$. If the inequality were strict, thenbecause the first factor is a nonzero nonnegative sum of simple roots and the second is strictly dominant. This contradicts the equal-norm condition. Only the orbit $W(\varphi +\rho )$ remains. Its leading coefficient is one, proving

For a finite group action on a finite set $X$, $R[X]$ is the free $R$-module with basis $X$, with the group permuting that basis. For a transitive set $G/H$ it is ${\mathrm{Ind}}_H^{G}R$. Disjoint unions of sets give direct sums of permutation modules.

For a real vector bundle $E$, its Pontryagin classes are the even-index Chern classes of its complexification with the displayed alternating signs. They lie in integral cohomology of degrees $4k$ and obey a total-class product formula under direct sums. If a metric connection on a vector bundle has curvature matrix $R$, its first real representative is $p_1(E)=-\mathrm{tr}(R\wedge R)/(8{\pi }^2)$. The Bianchi identity makes this representative closed, and Chern-Weil connection transgression makes its de Rham cohomology class independent of the connection. This normalization gives degree-four A-hat form $-p_1/24$.

For a linear map $T:V\to V$ on a finite-dimensional vector space over a field $F$, factor its minimal polynomial as $m_T={\prod }_i{p}_i^{a_i}$ with distinct monic polynomials $p_i$ that are irreducible polynomials. ThenEach $V_i$ is an invariant subspace, since $T$ commutes with every polynomial in $T$. These are canonical vector subspaces, determined by $T$, without requiring any eigenvalues in $F$.

For a proof, put $m_i=p_i^{a_i}$ and $M_i=m_T/m_i$. The coprime polynomials $m_i,M_i$ admit a Bézout's identity for polynomials $b_i{M}_i+c_i{m}_i=1$. Set $E_i=b_i(T)M_i(T)$. Its image is contained in $V_i$ because $m_i(T)E_i=0$. The polynomial ${\sum }_i{b}_i{M}_i-1$ is divisible by every $m_i$, hence by their product, so ${\sum }_i{E}_i=I$. On $V_i$, $E_i=I$; on $V_j$ with $j\ne i$, $E_i=0$ because $m_j$ divides $M_i$. These facts prove the direct sum and identify $E_i$ as its projection operators. Finally, if the minimal polynomial of any restriction had a smaller exponent, $m_T/p_i$ would annihilate every summand and hence all of $V$, a contradiction.

A linear projection is an idempotent linear operator: $P^2=P$. Every vector decomposes as $x=Px+(I-P)x$, into the image of a linear map and the kernel of a linear map. Those two vector subspaces form a direct sum. An orthogonal projection additionally uses the orthogonal complement in an inner product space.

A projective module is a direct summand of a free module. Free modules are flat because tensoring with a free module produces a direct sum of copies of the original module and map. If $F=P\oplus Q$ is flat, tensoring an injection with $F$ is the direct sum of its tensor maps with $P$ and $Q$, so each component is injective. Applying this to a free module containing $P$ proves that $P$ is flat.

Let $A_i$ be real symmetric matrices with sum $I_n$. Their matrix ranks sum to at least $n$, because their ranges span ${\mathbb{R}}^n$. If the ranks sum to $n$, those ranges form a direct sum. For $y$ in the range of $A_i$, the identity $y={\sum }_j{A}_{j}y$ and uniqueness of its range decomposition imply $A_{j}y={\delta }_{ij}y$. Thus $A_i^2=A_i$ and $A_i{A}_j=0$ for $i\ne j$: the matrices are mutually orthogonal orthogonal projections. In a combined orthonormal basis, the forms $Z^T{A}_{i}Z$ for $Z\sim N_n(0,I_n)$ use disjoint sets of independent standard normal coordinates. They are independent chi-squared distributions with degrees of freedom equal to the ranks. Conversely, if all these forms have those chi-squared laws, their expectations and their sum give ${\sum }_i\mathrm{rank}{A}_i=n$.

The representation ring is the Grothendieck group of finite-dimensional complex unitary representations of a compact group, with addition induced by direct sum and multiplication by tensor product. As an abelian group it is freely generated by the irreducible unitary representations. Taking dimension gives a ring homomorphism $R(G)\to \mathbb{Z}$.