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A standard Young tableau has its entries increasing from left to right in each row and from top to bottom in each column. For the convention $h_t=c_t{r}_t$, use column-reading order of standard Young tableaux: read columns from left to right, each from top to bottom, and compare the resulting words lexicographically. This makes the requested product direction explicit. The order and the symmetrizer multiplication convention must be chosen together.

We prove the vanishing claim. Suppose the rows of $t$ and the columns of $u$ have no collision. The argument in Question 1 shows that every column of $u$ contains one entry from each eligible row of $t$. Its first column thus selects one entry from every row. Each selected entry is at least the first entry of that row in $t$. Sorting the selected entries, as the standard tableau $u$ does, gives a column componentwise at least the first column of $t$: increasing individual entries cannot decrease any order statistic. If the columns are equal, equality of their sums forces every selected entry to be that row's first entry. Delete this common column and repeat. At the first differing column, its first differing entry in $u$ is therefore larger; otherwise $u=t$. Thus absence of a collision implies $u\ge t$ in column-reading order of standard Young tableaux.

If $t>u$, there must instead be a transposition in $R_t\cap C_u$. It fixes $r_t$ and negates $c_u$, giving $r_t{c}_u=0$. HenceThis is triangular vanishing of Young-symmetrizer products; it does not assert vanishing in the opposite order.

Normalize to $e_t=h_t/H_{\lambda }$. List all standard tableaux in increasing shape dictionary order on integer partitions, and within each shape in increasing column-reading order of standard Young tableaux. Then $e_i{e}_j=0$ for $i>j$, using Question 1 between different shapes and the result above within a shape. The left ideals $A{e}_i$ have an internal direct sum: if ${\sum }_i{z}_i=0$ with $z_i\in A{e}_i$, multiply on the right by $e_1$ to get $z_1=0$, since $z_1{e}_1=z_1$ and every later $z_i{e}_1=0$. Repeat with $e_2,e_3,\dots$. This proves directness without incorrectly treating all the idempotents as mutually orthogonal.

Let $f_{\lambda }$ count the standard tableaux of shape $\lambda$ and let $d_{\lambda }=\dim S^{\lambda }$. In the regular representation a simple module of dimension $d_{\lambda }$ occurs $d_{\lambda }$ times, by the Artin–Wedderburn theorem. The direct sum just constructed contains $f_{\lambda }$ copies of $S^{\lambda }$, so $f_{\lambda }\le d_{\lambda }$ for every shape. We supply the needed counting identity independently of the dimension conclusion.

The Robinson–Schensted correspondence bijects permutations with pairs of standard tableaux of the same shape. Here is its row insertion construction and inverse. Insert the successive permutation entries into an increasing row by replacing its first entry larger than the incoming entry, bumping that replaced entry into the next row; if no entry is larger, append at the row end. Continue until a new cell is created. Record the insertion time in that cell of a second tableau. For completeness, the successive bumped entries strictly increase, and their column indices weakly decrease: an entry below a bumped entry was originally larger, so the next replacement occurs no farther right. The entry newly placed in each row is smaller than the entry removed and larger than the entry above it. At a strictly earlier column, that last inequality follows from row increase in the preceding row; at the same column, it follows from the preceding bump. Hence the insertion tableau keeps increasing rows and columns. If a new cell is appended below the first row, the preceding bump guarantees that the row above reaches that column, so the shape stays a Young diagram. Recording times also increase in rows and columns, since every new cell is an outer corner of the current diagram. Thus both tableaux are standard at the end. Conversely, remove the cell with the largest recording label. Reverse its bumping path upward, replacing in each preceding row the rightmost entry smaller than the moving entry and moving the displaced entry upward. This recovers the last inserted letter; iterating recovers the entire permutation. The two rules undo one another at each row. ThusSemisimplicity also gives $n!={\sum }_{\lambda }{d}_{\lambda }^2$. Since $0\le f_{\lambda }\le d_{\lambda }$ termwise, equality of these sums forces $f_{\lambda }=d_{\lambda }$ for every shape. The internal direct sum has dimension $n!$ and hence fills $A$:There is also a useful numerical form. Right multiplication by $e_t$ is an idempotent with image $A{e}_t$. In the permutation basis of $A$, each diagonal coefficient is the coefficient of $1$ in $e_t$, namely $1/H_{\lambda }$. Its trace equals its rank, so $\dim S^{\lambda }=n!/H_{\lambda }$. Together with the count just obtained this recovers the hook-length formula.

The regression residual vector is $e=Y-X\hat{\beta }=(I-H)Y$, where $H=X(X^{T}X{)}^{-1}{X}^T$ is the hat matrix. Both $H$ and $I-H$ are symmetric idempotent matrices, and $(I-H)X=0$. Therefore $e=(I-H)\epsilon$ and the linear image of a multivariate normal vector givesThis multivariate normal distribution is singular, with rank $n-p$: fitted-residual orthogonality forces $X^{T}e=0$. Specifically, $\mathrm{Var}(e_i)={\sigma }^2(1-h_{ii})$ and $\mathrm{Cov}(e_i,e_j)=-{\sigma }^2{h}_{ij}$ for distinct observations. Thus regression residuals are generally neither independent nor identically distributed, even though the model errors are. Also $\mathrm{RSS}/{\sigma }^2\sim {\chi }_{n-p}^2$ and is independent of $\hat{\beta }$, which justifies the standard errors and exact Student t-tests below.

We take algebras to be unital and modules to be unital. A finite-dimensional semisimple algebra over the complex numbers is one whose regular left module is a semisimple module, that is, a direct sum of simple modules. We use the basic finite-dimensional equivalences: this is equivalent to a zero Jacobson radical, and the Jacobson radical of any finite-dimensional algebra is a nilpotent ideal.

The Artin–Wedderburn theorem here says that there are positive integers $d_1,\dots ,d_r$ withThe nonisomorphic simple modules are the standard column modules of the factors, of dimensions $d_i$, and the factor sizes are unique up to reordering.

Here is a proof. Decompose the regular left module as ${}_{A}A={⨁}_{i=1}^r{S}_i^{\oplus n_i}$ with pairwise nonisomorphic simple modules $S_i$. These are all the simple modules: any simple module is generated by a nonzero vector and is therefore a quotient of the regular module. The Schur lemma gives ${\mathrm{Hom}}_A(S_i,S_j)=0$ for $i\ne j$ and ${\mathrm{End}}_A(S_i)=\mathbb{C}$. For the latter assertion, an endomorphism has an eigenvalue over $\mathbb{C}$, and the kernel of its difference from that scalar is a nonzero submodule, hence the whole simple module. ConsequentlyEvery regular-module endomorphism is right multiplication by its value at $1$, so ${\mathrm{End}}_A({}_{A}A)\cong A^{\mathrm{op}}$. Taking opposite algebras and using matrix transposition gives the asserted decomposition with $d_i=n_i$.

For a matrix algebra $M_d(\mathbb{C})$, the matrix units show that its only simple module is ${\mathbb{C}}^d$: the spaces $E_{jj}S$ are isomorphic through $E_{ij}$, and a nonzero vector in one of them generates one copy of the column module. Simplicity makes that copy all of $S$. In a product algebra, the mutually orthogonal central idempotents decompose every module into its factor modules; a simple module uses exactly one factor. This proves the module assertion and also uniqueness, since the primitive central idempotents and the dimensions of their simple modules determine the factors.

For a Young tableau $t$, let $R_t$ permute the entries within its rows and $C_t$ within its columns. We use the Young symmetrizer conventionReversing the order gives another usual realization of the same irreducible polynomial module. On the tensor power $T=V^{\otimes n}$, the actions areThey commute because applying $g$ to every factor commutes with permuting the factors.

We state the permitted combinatorial input explicitly. A Young symmetrizer satisfies $h_t^2=H_{\lambda }{h}_t$, where $H_{\lambda }$ is the nonzero hook product of a partition. Thus $e_t=H_{\lambda }^{-1}{h}_t$ is a primitive idempotent. The standard-tableau decomposition of the right regular module is $\mathbb{C}{S}_n={⨁}_{\lambda ,t\mathrm{standard}}{e}_t\mathbb{C}{S}_n$. Tensor this right-module direct sum with the left module $T$. The map $e_t\mathbb{C}{S}_n{\otimes }_{\mathbb{C}{S}_n}T\to e_{t}T$ sending $e_{t}a\otimes v$ to $e_{t}av$ is an isomorphism, with inverse $w\mapsto e_t\otimes w$. Therefore the Young-symmetrizer tensor decomposition isThis is a direct sum of $GL(V)$-modules; individual summands need not be $S_n$-invariant.

We also state the allowed Schur algebra result, namely Schur–Weyl duality: the two actions are mutual commutants, andwhere the $D_{\lambda }(V)$ are pairwise nonisomorphic irreducible homogeneous polynomial representations of degree $n$. We also use the standard Schur algebra equivalence between its modules and homogeneous degree-$n$ polynomial representations, so these exhaust the irreducibles in that category. A primitive idempotent $e_t$ has one-dimensional image on $S^{\lambda }$ and zero image on the other simple factors, so $h_{t}T\cong D_{\lambda }(V)$. This identifies the requested irreducibles. They classify the polynomial degree-$n$ representations in this tensor power, not all rational representations of every degree.

The length bound for a Schur module is $D_{\lambda }(V)\ne 0$ if and only if $\ell (\lambda )\le m$. A column of length greater than $m$ antisymmetrizes more than $m$ vectors and gives zero. Conversely, for $\ell (\lambda )\le m$, fill every tensor position in row $i$ with the $i$th basis vector of $V$. Row symmetrization multiplies it by ${\prod }_i{\lambda }_i!$, and column antisymmetrization is nonzero because the vectors within each column are distinct basis vectors.

A rational representation of an algebraic group is a regular morphism into the general linear group of its representation space. For $G{L}_m$, its matrix entries belong to $\mathbb{C}[g_{ij},\det (g{)}^{-1}]$; rational here permits determinant denominators but not arbitrary poles on $G{L}_m$. A one-dimensional rational character of ${\mathbb{C}}^{\times }$ is a Laurent polynomial $p(z)$ with $p(zw)=p(z)p(w)$ and $p(1)=1$. Comparing Laurent coefficients shows that just one monomial occurs and its coefficient is $1$, hence $p(z)=z^r$ for $r\in \mathbb{Z}$.

Restrict a one-dimensional rational character of $G{L}_m$ to its diagonal torus. The same argument in several variables gives ${\prod }_i{x}_i^{r_i}$. Conjugation by permutation matrices makes all $r_i$ equal. On every diagonalizable invertible matrix it consequently agrees with $(\det g{)}^r$. The allowed Zariski-density statement, and equality of regular functions on a dense subset, giveThese are the one-dimensional rational characters of the general linear group.

For complete reducibility of rational GL and SL representations, use the compact-group averaging argument. Average any positive definite Hermitian inner product over $SU(m)$ using normalized Haar measure. The orthogonal complement of an invariant subspace is then $SU(m)$-invariant. Differentiating makes it invariant under $\mathfrak{su}(m)$ and therefore under its complex span ${\mathfrak{sl}}_m(\mathbb{C})$. The elementary unipotent matrices $\exp (t{E}_{ij})$ generate $S{L}_m(\mathbb{C})$, so the complement is $S{L}_m$-invariant. This proves complete reducibility. Averaging over $U(m)$ similarly gives complete reducibility for rational $G{L}_m$ representations.

Here is an explicit rational extension from SL to GL. Decompose the representation space by the finite scalar center of $S{L}_m$ into subspaces $W_k$ on which $\zeta I$ acts as ${\zeta }^k$, with $0\le k<m$ and ${\zeta }^m=1$. These subspaces are invariant. For $g\in G{L}_m$, choose $a$ with $a^m=\det g$, set $s=a^{-1}g\in S{L}_m$, and defineChanging $a$ to $\zeta a$ changes $s$ to ${\zeta }^{-1}s$, so the two factors cancel. It is a homomorphism, since scalar roots multiply up to the same harmless factor, and it restricts to $\rho$ on $S{L}_m$.

It is rational as well. Every matrix coefficient of $\rho {|}_{W_k}$ has a polynomial representative on $S{L}_m$. Averaging that representative over the finite scalar center selects its homogeneous parts $P_d$ with $d\equiv k(\mathrm{mod}m)$. Substitution in the extension gives ${\sum }_d(\det g{)}^{(k-d)/m}{P}_d(g)$, a regular function on $G{L}_m$. Each $S{L}_m$-invariant subspace decomposes into its $W_k$ parts, and the extension acts on each part by a scalar times an $S{L}_m$ action. Thus these subspaces are also invariant under the chosen extension. Consequently $\rho$ is irreducible if and only if this ${\rho }^{\prime }$ is irreducible. For $m=1$, $S{L}_1$ is trivial and the trivial extension supplies the same conclusion.

For a finite acyclic quiver, the arrow ideal of a path algebra $R$ is nilpotent, and $A/R\cong {\prod }_{i\in Q_0}k$. If $S$ is a simple module, its submodule $RS$ is either zero or $S$. The latter would imply $R^{d}S=S$ for every $d$, contradicting nilpotence. Thus $RS=0$, and a simple module over the product of fields is supported at one coordinate. Therefore the simples are exactly $S(i)$, with $k$ at $i$, zero elsewhere, and zero arrows; the vertex $i$ is unique.

A finite-dimensional semisimple module is consequently ${⨁}_{i}S(i{)}^{\oplus n_i}$, where $n_i=\dim X_i$. Its dimension vector of a quiver representation determines its isomorphism class.

For an arbitrary finite quiver, cycles allowed, the vertex projective module of a path algebra is $P(i)=A{e}_i$. Its space at vertex $j$ has basis all paths from $i$ to $j$, and an arrow acts by adjoining that arrow at the end of the path. Its endomorphism ring iswhere $e_{i}A{e}_i$ is spanned by the closed paths based at $i$. The opposite multiplication appears because endomorphisms act by right multiplication.

The evaluation isomorphism for a vertex projective isFor $x\in X_i$, its inverse sends a path $p$ starting at $i$ to $px$. This proves both injectivity and surjectivity, and is natural in $X$. Vertex evaluation is exact, so $P(i)$ is a projective module; alternatively it is a direct summand of the free module $A$.

The closed-path corner of a path algebra is a domain: in a product of two nonzero linear combinations, choose their longest path lengths. Concatenation in that top degree has a unique cut at those lengths, so a product of two nonzero top coefficients cannot cancel. Hence its only idempotents are zero and one. The same is true of the opposite ring, proving $P(i)$ is an indecomposable module, even when cycles make it infinite-dimensional.

For $1\to 2\leftarrow 3$, the paths starting at vertex $1$ are $e_1$ and the arrow $1\to 2$. Thus the displayed $k\stackrel{1}{\to }k\leftarrow 0$ is $P(1)$ and is projective.

A nonzero finite-dimensional representation is a brick module when its endomorphism ring is a division algebra. Over the algebraically closed field $k$, this means ${\mathrm{End}}_Q(X)=k$: for any endomorphism $f$, an eigenvalue $\lambda$ makes $f-\lambda I$ noninvertible, hence zero in a division algebra.

For a counterexample to the converse of “brick implies indecomposable”, take the one-loop representation $k^2$ with nilpotent Jordan block $N=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$. Its endomorphism ring is $k[N]\cong k[t]/(t^2)$, a local endomorphism ring of dimension two. It has no nontrivial idempotents, so the module is indecomposable, but it is not a brick.

For the one-arrow quiver, splitting the kernel, image and target complement decomposes any representation into copies of $k\to 0$, $0\to k$ and $k\stackrel{1}{\to }k$. Each has endomorphism ring $k$. Therefore every indecomposable of the one-arrow quiver is a brick.

Write $h_{ij}={}^{(3)}{g}_{ij}$ for the positive spatial metric tensor and $h^{ij}$ for its inverse. Expanding the 3+1 decomposition gives $g_{00}=N^2-h_{ij}{N}^i{N}^j$, $g_{0i}=-h_{ij}{N}^j$ and $g_{ij}=-h_{ij}$. The inverse isFor example, $g^{00}{g}_{00}+g^{0i}{g}_{i0}=1$, and $g^{00}{g}_{0j}+g^{0i}{g}_{ij}=0$. The spatial block similarly gives $g^{i0}{g}_{0j}+g^{ik}{g}_{kj}={\delta }^i{}_j$. These checks determine all blocks without treating the spatial block alone as the inverse of the four-metric.

Raising the normal covector with this inverse metric tensor yieldsTake $N>0$ so it is future-pointing. Its norm is $n_{\mu }{n}^{\mu }=1$. The spatial projection tensor obeys $P^{\mu }{}_{\nu }{n}^{\nu }=0$, and multiplication givesThus it is an idempotent linear projection onto vectors tangent to the spatial hypersurface. The negative spatial components of the spacetime metric tensor do not change this idempotence.

The path algebra $kQ$ is the $k$-vector space with basis all directed paths, including the length-zero path $e_i$ at each vertex. Extend path composition bilinearly, declaring the product zero when the paths cannot be composed. We use the left-module convention: $pq$ means first traverse $q$, then $p$; for an arrow $\rho :i\to j$, $e_j\rho e_i=\rho$.

The constant paths are mutually orthogonal idempotents. For any path $p:i\to j$,Consequently, for the finite vertex set used here,It acts as an identity on every basis path, hence on the entire path algebra. Finiteness of the vertex set is what makes this a genuine element of the algebra; an infinite-vertex path algebra instead has finite vertex sums as local units.

Number the source vertices $1,\dots ,r$ and the sink $c=r+1$, and write $a_i:i\to c$ for the arrows. The only basis paths are $e_1,\dots ,e_r,e_c,a_1,\dots ,a_r$. Using standard matrix units, defineThe relations $e_c{a}_i=a_i=a_i{e}_i$, orthogonality of the vertex idempotents, and $a_i{a}_j=0$ agree exactly with multiplication of these matrix units. The map is therefore an algebra homomorphism, and its images form a basis of the algebraThis is an explicit isomorphism of the path algebra of the subspace quiver with a $(2r+1)$-dimensional triangular matrix algebra.

Set $A=kQ$, $X_i=e_{i}X$, and let $f_{\rho }$ be the action of the arrow $\rho$ on $X_{s(\rho )}$. Write $P_i=A{e}_i$, a left projective module. The standard projective resolution of a quiver representation isThe algebra acts on the first tensor factor. The augmentation is $\epsilon (p\otimes x)=px$. On the summand for $\rho :i\to j$, the differential sends $p\otimes x$ to $p\rho \otimes x$ in the $i$-summand minus $p\otimes f_{\rho }(x)$ in the $j$-summand. Here $p\in A{e}_j$, so $p\rho \in A{e}_i$. These formulas specify every term and map, and $\epsilon d=0$.

Exactness is the standard path resolution fact: the relations identify a path acting on a vector with successively applying its arrows; uniqueness of the first traversed arrow supplies injectivity of the relation map. Each $A{e}_i$ is a direct summand of $A$ because $e_i$ is an idempotent, and tensoring with a $k$-vector space gives a direct sum of copies of $A{e}_i$. Both terms preceding $X$ are consequently projective modules. Thus the displayed sequence is a projective resolution of length at most one.

The same path resolution works for arbitrary left modules, with possibly infinite-dimensional $X_i$ and arbitrary direct sums of projective modules. Hence every left $A$-module has projective dimension at most one. Equivalently, $A$ is a left hereditary ring: if $N\subseteq P$ with $P$ projective, dimension shifting gives ${\mathrm{Ext}}_A^1(N,M)\cong {\mathrm{Ext}}_A^2(P/N,M)=0$ for every $M$, so $N$ is projective. This argument also covers quivers with oriented cycles.

For dimension vectors $\mathbf{n},\mathbf{m}\in {\mathbb{R}}^{Q_0}$, define the Ringel formApply ${\mathrm{Hom}}_A(-,Y)$ to the projective resolution. Evaluation at $e_i$ identifies ${\mathrm{Hom}}_A(P_i{\otimes }_{k}V,Y)$ with ${\mathrm{Hom}}_k(V,Y_i)$. Consequently there is an exact sequencewhereThis is the extension complex of quiver representations. Taking its alternating dimension sum yields

A nonzero finite-dimensional quiver representation is a brick module when ${\mathrm{End}}_Q(X)=k{1}_X$. If $X=U\oplus V$ with both summands nonzero, projection onto $U$ is an idempotent endomorphism different from zero and the identity. That is impossible in $k{1}_X$. Hence every brick is indecomposable.

For the requested converse counterexample, take the dual-number algebra $A=k[\epsilon ]/({\epsilon }^2)$ and its left regular module $A$. The algebra is local with maximal ideal $(\epsilon )$, and it is not simple because this is a nonzero proper ideal. The module is indecomposable: its endomorphism ring is $A^{\mathrm{op}}\cong A$, which has no nontrivial idempotents. But this ring has dimension two over $k$, so the regular module is not a brick. Thus indecomposable need not mean brick without the positive-definiteness hypothesis used below.