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Let $P$ be the averaging linear mapFor a unitary representation of a finite group, $P$ is the orthogonal projection onto the invariant subspace $V^G$, and $P\rho (h)=\rho (h)P=P$. Applying this to $\rho (g)v=\lambda v$ gives $Pv=\lambda Pv$. Since $\lambda \ne 1$, $Pv=0$. The vector $v$ is a cyclic vector for a group representation, so its translates span $V$, and every translate also has projection zero. Thus $P=0$ andThis is the cyclic eigenvector obstruction to invariant vectors. Unitarity also gives a direct proof by pairing $v$ with an invariant vector and then pairing every translate; the averaging proof makes the role of cyclicity especially transparent.

It suffices to handle one single-box up-move, and then use the chain already constructed. Suppose the affected row sizes are $a={\lambda }_i$ and $b={\lambda }_j$, $i<j$. Since $\lambda$ is a partition of an integer, $a\ge b\ge 1$. Put $N=a+b$. Over the complex numbers, the two-row Young permutation module decomposition givesA zero second row is omitted. Therefore

Let $K$ be the product of the symmetric groups of the unaffected rows and the symmetric group $S_N$ on the union of the affected rows. The two relevant Young subgroups lie in $K$. Transitivity of induced representations expresses $M^{\lambda }$ and $M^{\nu }$ by inducing the preceding two-row modules, tensored with the trivial representations of the unaffected factors, from $K$ to $S_n$. Induction preserves this direct sum, sofor an actual group representation $L_{\lambda ,\nu }$, not merely a difference of characters. Iterate along the chain and take the direct sum of the induced complements. The resulting complement can be realized as an invariant subspace of $M^{\lambda }$, either through these isomorphisms or by Maschke's theorem. ThusIf $\lambda =\mu$, use the zero complement. The characteristic-zero hypothesis is essential to this use of the irreducible two-row decomposition.

Use the nonnegative Hodge-Laplacian convention from Question 4. The Bochner-Weitzenbock formula for one-forms isHere $\nabla$ is the Levi-Civita connection induced on the cotangent bundle, and the rough Laplacian in a local orthonormal frame isIt is the composition of covariant derivative with its formal adjoint. The Ricci endomorphism is defined by $g({\mathrm{Ric}}^{♯}X,Y)=\mathrm{Ric}(X,Y)$ and acts on one-forms by $(\mathrm{Ric}\cdot \alpha )(X)=\alpha ({\mathrm{Ric}}^{♯}X)$. The musical isomorphism defines ${\alpha }^{♯}$ by $g({\alpha }^{♯},X)=\alpha (X)$. The associated scalar formula isThese signs make the integrated rough-Laplacian term $\Vert \nabla \alpha {\Vert }^2$ on a closed manifold.

For a connected manifold, the full Riemannian holonomy group at $p$ is the subgroup of $O(T_{p}M,g_p)$ consisting of parallel transports around all piecewise smooth loops based at $p$. Its natural action on $T_{p}M$ is the holonomy representation; it induces actions on cotangent spaces and all tensor spaces. The holonomy representation is an irreducible representation when it has no nonzero proper invariant subspace.

The fundamental principle of Riemannian holonomy identifies parallel tensor fields with tensors at $p$ fixed by full holonomy. A parallel field returns to its value under every loop. Conversely, transport a fixed tensor along a path from $p$ to each point. Any two paths differ by a loop, so the result is independent of the path; local smooth parallel transport yields a smooth parallel field. Evaluation and construction are inverse. Full holonomy, not merely the contractible-loop subgroup, is required for this global correspondence.

On a compact manifold without boundary and with nonnegative Ricci curvature, a harmonic one-form satisfies the integrated Bochner identityBoth terms are nonnegative; therefore $\nabla \alpha =0$. This proves that harmonic one-forms are parallel under nonnegative Ricci curvature. Integration uses the Riemannian volume density and does not require an orientation.

If $n=\dim M\ge 2$, a nonzero such form would give a nonzero fixed tangent vector via metric duality and hence a proper invariant line, contradicting irreducibility. This is precisely the irreducible holonomy in dimension at least two has no parallel one-form criterion. Since the question permits harmonic representatives of all classes, it gives $b_1(M)=0$, where $b_1$ is the first Betti number.

The covering $\pi :X\times T^k\to M$ is finite: its fibre over a point is a closed discrete subset of the compact total space and hence finite. Use the explicitly permitted equality of the Betti numbers of $M$ and its finite cover, in particular $b_1(X\times T^k)=b_1(M)=0$. This equality is a permission of this question, not a general theorem about finite covers.

The $k$ angular one-forms on $T^k$, pulled back to $X\times T^k$, represent linearly independent de Rham classes: their periods on the $k$ coordinate circles are the standard basis vectors. exact differential forms have zero periods. Thus $b_1(X\times T^k)\ge k$, forcing $k=0$. Now $\pi :X\to M$ is a finite universal covering map because $X$ is simply connected. The fundamental group acts freely and transitively on a fibre, or equivalently loop-lifting identifies its elements with the finite set of possible endpoints upstairs. Consequently the intended conclusion isThe printed statement needs the dimension qualification under the usual definition of irreducibility. Take the standard circle $M=S^1$, with $X$ a point, $k=1$, and the identity covering $X\times T^1\to S^1$. Its Ricci curvature is zero. parallel transport fixes its global unit tangent, so its holonomy representation is the trivial representation on a one-dimensional real vector space, which is irreducible. The allowed Betti-number equality holds for this identity cover, yetThus there is no proof of the unqualified literal assertion in dimension one. If “irreducible holonomy” is instead intended to exclude the one-dimensional trivial representation, the preceding intended proof applies. A connected zero-dimensional manifold is a point and has trivial fundamental group.

Assume first $0<k<N$. Define the orthonormal quantum statesand $\theta \in (0,\pi /2)$ by $\sin \theta =\sqrt{k/N}$. The relevant invariant subspace is the two-dimensional span of these quantum states, not the entire Hilbert space. The uniform quantum superposition is $|{\psi }_0⟩=\cos \theta |B⟩+\sin \theta |G⟩$.

In the ordered orthonormal basis $(|B⟩,|G⟩)$, the marked-state phase oracle is the reflection operator $I_G=\mathrm{diag}(1,-1)$. Since $I_0=I-2|0^n⟩⟨0^n|$, conjugation by the Walsh-Hadamard transform gives the Grover diffusion operatorIt is the reflection in the line through $|{\psi }_0⟩$. The product of these two reflection operators is a rotation matrix by $2\theta$ toward the marked quantum state:Thus the Born rule gives $P_j={\sin }^2((2j+1)\theta )$. Choose a nonnegative integer $j_{\ast }$ nearest to $\pi /(4\theta )-1/2$. Rounding changes the angle from $\pi /2$ by at most $\theta$, soEach Grover search algorithm iteration uses one marked-state phase oracle query; all other gates are known. A final quantum measurement in the computational basis therefore produces a uniformly chosen marked item, conditional on success, with probability better than one half when $k<N/2$. This uses the supplied marked count $k$ to select the stopping time; an unknown $k$ needs an additional search schedule or counting procedure. If $k=N$, no oracle query is needed; if $k=0$, no marked item exists, so a successful search is impossible.

Give $V_n$ the action $(g\cdot P)(x,y)=P((x,y)g)$, where $(x,y)$ is a row vector. Substitution preserves homogeneous degree, and composition agrees with matrix multiplication, so this is a complex representation of SU(2) group. The diagonal torus $\mathrm{diag}(z,z^{-1})$ acts on $x^{n-j}{y}^j$ with weight $z^{n-2j}$; these weights are distinct.

A nonzero invariant subspace is invariant under the torus, so finite-dimensional Fourier projection onto its weight spaces supplies a nonzero monomial. Differentiating the group action and complexifying supplies the operators $E=x{\partial }_y$ and $F=y{\partial }_x$. Their repeated application moves between every successive monomial, with nonzero coefficients until an endpoint is reached:The invariant subspace therefore contains the whole monomial basis. ThusFor $n=0$ this is the one-dimensional trivial representation. Torus weight projections are legitimate because an invariant complex subspace is closed in finite dimension.

A two-dimensional Lie algebra representation already provides a counterexample. In the basis $e_1,e_2$ of $V={\mathbb{C}}^2$, defineThe commutator $[\rho (x),\rho (y)]$ is zero, agreeing with $\rho (z)$; the other defining Lie brackets also map to zero. Thus $\rho$ is a Lie algebra homomorphism, although it is not a Faithful Lie algebra representation.

The line $L=\mathbb{C}{e}_1$ is an invariant subspace. Any complementary line has the form $\mathbb{C}(e_2+a{e}_1)$ for some $a\in \mathbb{C}$. Butso no complementary line is an invariant subspace. Equivalently, every invariant line is killed by $\rho (x)$: its restriction to a line is a scalar, and its square is zero, forcing that scalar to be zero. A decomposition into two invariant lines would then force $\rho (x)=0$, a contradiction. This representation is therefore not a semisimple representation. Thus

If the spectrum has one point, the preceding part gives $T=\lambda I$, and the linear span of any nonzero vector is a closed proper invariant subspace because $\dim H>1$.

Otherwise choose distinct points ${\lambda }_1,{\lambda }_2\in K$ and disjoint nonempty relatively open neighborhoods $U,V\subseteq K$. The open-set result gives $P(U)\ne 0$ and $P(V)\ne 0$, whileThus $P(U)\ne I$. Set $L=\mathrm{ran}P(U)$. As the range of an orthogonal projection, $L$ is closed, and the preceding observations give $\{0\}\ne L\ne H$. The Borel functional calculus for a normal operator gives $TP(U)=P(U)T$, soIn fact both $L$ and its orthogonal complement are invariant.

Let $W=\mathrm{span}\{\mathbf{v},\mathbf{w}\}=\mathrm{col}V$. Since the two columns are linearly independent, the leading $2\times 2$ block $R_1$ inis invertible. Hence $S(W)=\mathrm{span}\{{\mathbf{e}}^{(1)},{\mathbf{e}}^{(2)}\}$. The hypothesis that $W$ is an invariant subspace of $A$ implies that this coordinate plane is invariant under $\hat{A}=SA{S}^{-1}$. Consequently the first two columns of $\hat{A}$ have no entries below row two, andwhere $B$ is the top-left $2\times 2$ block and $D$ is the bottom-right $(n-2)\times (n-2)$ block. Eigenvalue deflation by an invariant subspace now givessoagain counting algebraic multiplicities.

For the required orthogonal reduction, take a thin QR decomposition $V=Q_1{R}_1$, extend the two orthonormal columns of $Q_1$ to an orthogonal matrix $Q$, and set $S=Q^T$. ThenEquivalently, apply one Householder transformation to annihilate entries $2,\dots ,n$ of $\mathbf{v}$, followed by a second Householder transformation acting only on coordinates $2,\dots ,n$ to annihilate entries $3,\dots ,n$ of the transformed $\mathbf{w}$.

Schur lemma over $\mathbb{C}$ states that an intertwining operator $T:V\to W$ between finite-dimensional irreducible complex representations is either zero or an isomorphism. In particular, every intertwining endomorphism of $V$ is a scalar multiple of the identity.

Indeed, $\ker T\subseteq V$ and $\mathrm{im}T\subseteq W$ are invariant subspaces. If $T\ne 0$, irreducibility gives $\ker T=0$ and $\mathrm{im}T=W$, so $T$ is an isomorphism. For an endomorphism $T:V\to V$, the fundamental theorem of algebra gives an eigenvalue $\lambda \in \mathbb{C}$. Then $T-\lambda I$ is an intertwiner with nonzero kernel, so the first part forces $T-\lambda I=0$. Hence $T=\lambda I$, completing the Proof of Schur lemma.

Maschke's theorem says that if $H$ is a finite group and the characteristic of a field $F$ does not divide $|H|$, then every finite-dimensional $F$-representation of $H$ is semisimple.

To prove it, let $W$ be an invariant subspace of a representation $V$, and choose any linear projection $P:V\to W$. Average its conjugates:The scalar $|H{|}^{-1}$ exists by the characteristic assumption. Reindexing the sum shows that $\overline{P}$ commutes with every $\rho (g)$. Since each summand restricts to the identity on $W$, so does $\overline{P}$; hence $\overline{P}$ is an equivariant projection onto $W$. Its kernel is an invariant complement, and induction on $\dim V$ decomposes $V$ into irreducible representations.

The isometry group of $\mathbb{Z}$ is the infinite dihedral groupLet $\rho$ be a nonfaithful finite-dimensional complex representation. Its nontrivial normal kernel contains $t^m$ for some $m>0$, by the stated hint, so $\rho$ factors through the finite quotientMaschke's theorem decomposes this quotient representation into irreducibles. To bound their dimensions, choose a $t$-eigenvector $v$ with eigenvalue $\lambda$ in an irreducible constituent. The relation givesso $\mathrm{span}\{v,sv\}$ is invariant under both generators. Irreducibility makes it the whole constituent, whose dimension is therefore at most two. This proves the result on nonfaithful representations of the infinite dihedral group.

For the requested representation of $(\mathbb{Z},+)$, let its generator act on the polynomials of degree at most one by translation:In the basis $(1,x)$,This is a nontrivial Jordan block, so it is not diagonalizable. A direct sum of one-dimensional subrepresentations would make $T$ diagonalizable, proving that this nonsemisimple translation representation of the infinite cyclic group has the required property.

Finally, on $V=\mathbb{C}[x{]}_{\le 2}$ defineThen $S^2=I$ and $STS=T^{-1}$, so these operators define a quadratic-polynomial representation of the infinite dihedral group. In the basis $(1,x,x^2)$ they areHere $(T-I{)}^2\ne 0$ but $(T-I{)}^3=0$, so $T$ consists of one size-three Jordan block. If $V$ were a direct sum of subrepresentations of dimensions at most two, then $T$ would be block diagonal with Jordan blocks of sizes at most two, a contradiction. Thus this representation has the required failure of decomposition.

For $y\in D(A)$, the semigroup property givesHence $U(t)y\in D(A)$ andThe generator domain is therefore an invariant subspace, and the operator norm bound givesThusMoreover, strong continuity applied separately to $y$ and $Ay$ givesTherefore the restrictions form the semigroup restricted to its generator domain. Its derivative at zero exists in the graph norm exactly when $y\in D(A)$ and $Ay\in D(A)$, namely when $y\in D(A^2)$, and then the derivative is $Ay$. Hence its generator is

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.

The two nonzero proper invariant subspaces areBoth are preserved by the rotation. To show there are no others, note thatis the linear projection onto $U$, and $I-P$ is the linear projection onto $W$. Every invariant subspace $L$ is preserved by $I,T^2$, and hence by both of these linear projections. ThereforeBecause $U$ has dimension one, $L\cap U$ is either $0$ or $U$. If $L\cap W$ contains a nonzero $v=(a,b,0)$, it also contains $Tv=(-b,a,0)$. These two vectors are linearly independent, because their two-dimensional determinant is $a^2+b^2>0$. They span $W$, so $L\cap W$ is either $0$ or $W$. The only possibilities for $L$ are therefore $0,U,W,V$.

Finally every $(x,y,z)$ has the unique decomposition $(0,0,z)+(x,y,0)$, with one summand in each of $U,W$. HenceThis also illustrates how invariant subspaces of a quarter-turn in three dimensions can be classified by linear projections.

The real spectral theorem states that a symmetric matrix has an orthonormal basis of real eigenvectors, and hence is a diagonalisable matrix. Here is a proof in this setting.

For $n\ge 1$, the quadratic form $q(x)=x^{T}Ax$ attains a maximum on the unit sphere, by compactness and the extreme value theorem. Choose a unit maximizer $u$ and put $\lambda =u^{T}Au$. For any $w\perp u$, the Rayleigh quotient along $u+tw$ has a maximum at $t=0$, soThe last equality uses $A=A^T$ and $u^{T}w=0$. Thus $Au$ is orthogonal to every $w\perp u$, so $Au=\lambda u$.

The orthogonal complement $u^{\perp }$ is an invariant subspace, because for $x\perp u$,The restriction of $A$ to $u^{\perp }$ is again symmetric with respect to the induced inner product. By mathematical induction on dimension, it has an orthonormal basis of eigenvectors; adjoining $u$ gives one for ${\mathbb{R}}^n$. If $Q$ has those basis vectors as columns, then $Q^{T}AQ$ is a real diagonal matrix and $Q^{-1}=Q^T$.

A group representation is a homomorphism $D:G\to \mathrm{GL}(V)$. It is a reducible representation if there is a nonzero proper invariant subspace of $V$ preserved by every $D(g)$, and an irreducible representation otherwise. With $S$ invertible,so changing basis gives another representation. Its character of a representation is $\chi (g)=\mathrm{tr}D(g)$, and cyclicity of the trace gives $\tilde{\chi }(g)=\mathrm{tr}(SD(g)S^{-1})=\chi (g)$.

A conjugacy class consists of $hg{h}^{-1}$ as $h$ varies. Since $(hg{h}^{-1}{)}^k=h{g}^k{h}^{-1}$, a power equals the identity exactly when $g^k$ does. Thus conjugation preserves element order, including infinite order.

For the square, let $r$ be rotation by $\pi /2$ and $s$ reflection in a horizontal symmetry axis. The dihedral group has the eight elementsThe conjugacy classes of a finite dihedral group specialize toThe two central rotations are singletons. Reflection conjugation reverses rotations, while conjugation by $r$ shifts a reflection exponent by two, giving the two reflection classes.

For finite groups over $\mathbb{C}$, character orthogonality and completeness make the irreducible characters a basis of the class functions. The latter space has one independent value per conjugacy class. Hence the square group has five inequivalent irreducible complex representations. The sum of squares of irreducible representation dimensions is eight, forcing four one-dimensional representations and one two-dimensional representation. The four scalar choices $r\mapsto \pm 1$, $s\mapsto \pm 1$ supply the former; the ordinary action on the plane supplies the latter.

An invariant subspace $W$ satisfies $D(g)W\subseteq W$ for every $g\in G$; applying the inverse gives equality. The representation is irreducible when the only invariant subspaces are $0$ and the full space.

The inverse and conjugate transpose each reverse multiplication order, so their combination preserves it:Its values are invertible, so this is the conjugate contragredient representation.

If $W$ is invariant under $D_2$, take $w\in W$ and $v\in W^{\perp }$. For the standard Hermitian inner product,since $D_2(g^{-1})w\in W$. Hence $D_1(g)v\in W^{\perp }$, proving its invariance. If $D_1$ is irreducible, $W^{\perp }$ is either zero or the whole space; finite-dimensional orthogonal complementation makes $W$ the whole space or zero, respectively. Thus $D_2$ is irreducible whenever $D_1$ is irreducible.

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.

Denote the four listed matrices, in order, by $I,P,Q,W$. As permutation matrices, $P,Q,W$ act on the four basis vectors by $(12)(34)$, $(13)(24)$ and $(14)(23)$, respectively. Direct multiplication givesThese relations prove closure and show that each element is its own inverse. The identity is included and matrix multiplication is associative, so the matrices form a group.

The relations coincide with those of the Klein four-group $D_2$: take its half-turn generator to $P$, a reflection generator to $Q$, and their product to $W$. All four matrices are distinct, so this assignment is a faithful representation of $D_2$. It is the four-point regular representation of the Klein four-group.

Every matrix fixes $v=(1,1,1,1{)}^T$, since it permutes its coordinates. Thereforeis a one-dimensional invariant subspace, nonzero and strictly smaller than the four-dimensional space. This proves the representation is reducible. In fact the four mutually orthogonal sign vectors $(1,1,1,1)$, $(1,1,-1,-1)$, $(1,-1,1,-1)$ and $(1,-1,-1,1)$ give invariant lines and completely diagonalize the representation, as in the normalized Hadamard matrix construction.

For a group representation $D:G\to \mathrm{GL}(V)$, a faithful representation is injective, equivalently $D(g)=I$ only for $g=e$. Thus no distinct group elements are represented by the same linear map.

Two equivalent representations are related by one fixed invertible change of basis: $E(g)=S^{-1}D(g)S$ for every $g$. The same $S$ must work simultaneously for all group elements.

An irreducible representation acts on a nonzero vector space and has no invariant subspace other than $\{0\}$ and the whole space. An invariant subspace is one carried into itself by every representing matrix.

The character of a representation isThe cyclic property of the trace makes it invariant under equivalence and constant on conjugacy classes. Its value at the identity is the dimension of the representation.

A vector subspace $W\subseteq {\mathbb{C}}^n$ is an invariant subspace for the group representation $D$ ifBecause $g^{-1}$ is also in the group and $D(g)$ is invertible, the inclusion is equivalently $D(g)W=W$ for every $g$.

For $n>0$, the representation is an irreducible representation when its only invariant vector subspaces areThus no nonzero proper subspace is preserved by all representing matrices.