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Let $N=|R^{+}|$ and identify the real weight and Cartan spaces using a Weyl group-invariant positive inner product. For a regular $H$, meaning $(\alpha ,H)\ne 0$ for every root, specialize $e^{\mu }$ to $e^{t(\mu ,H)}$. Then the value of the formal character at $t=0$ is $\dim L(\lambda )$, but both Weyl alternants in the Weyl character formula vanish there. We compute their leading terms instead of substituting into that zero-over-zero expression.

ExpandFor every root reflection $s$, replacing $\nu$ by $s\nu$ changes the sign of $P_m$, by replacing $w$ with $ws$ in the sum. Replacing $H$ by $sH$ also changes its sign, by replacing $w$ with $sw$. Thus the polynomial vanishes whenever $\nu$ or $H$ lies on any reflecting hyperplane. It is divisible by ${\prod }_{\alpha >0}(\nu ,\alpha )$ in its first variables and by ${\prod }_{\alpha >0}(H,\alpha )$ in its second variables. Distinct positive roots give distinct linear factors; each product has degree $N$.

The polynomial $P_m$ has degree $m$ in each group of variables. Therefore $P_m=0$ for $m<N$, and for $m=N$ it has the formwith a constant $C$ independent of both variables. At $\nu =\rho$, the Weyl denominator formula givesConsequently $C{\prod }_{\alpha >0}(\rho ,\alpha )=1$. For strictly dominant $\nu =\lambda +\rho$, its leading coefficient is nonzero, and the ratio of the two Weyl alternants tends toThe coroot is ${\alpha }^{\vee }=2\alpha /(\alpha ,\alpha )$, so the root-length factors cancel in each ratio. This proves the Weyl dimension formula by the lowest-degree alternant proof of the Weyl dimension formula, including the limiting step.

Give the real root span an inner product with ${\epsilon }_1,{\epsilon }_2$ orthonormal. The B2 root system consists of four short roots on the coordinate axes and four long roots on the diagonals. Choose the long-first simple rootsThe positive roots are ${\alpha }_1,{\alpha }_2,{\alpha }_1+{\alpha }_2={\epsilon }_1$, and ${\alpha }_1+2{\alpha }_2={\epsilon }_1+{\epsilon }_2$. The corresponding simple reflections areThey generate every signed permutation of the two coordinates. Thus the Weyl group has order eight and is the symmetry group of a square; every element sends $(x,y)$ to $(\pm x,\pm y)$ or $(\pm y,\pm x)$, with independent signs.

Since ${\alpha }_1,{\alpha }_2$ form an integral basis of ${\mathbb{Z}}^2$, the root lattice is $Q=\mathbb{Z}{\epsilon }_1\oplus \mathbb{Z}{\epsilon }_2$. The simple coroots are ${\alpha }_1^{\vee }={\epsilon }_1-{\epsilon }_2$ and ${\alpha }_2^{\vee }=2{\epsilon }_2$. Therefore the weight lattice isSolving $⟨{\Lambda }_i,{\alpha }_j^{\vee }⟩={\delta }_{ij}$ gives the fundamental weightsThe cone of dominant weights isIndeed $a=x-y$ and $b=2y$ are exactly the two simple-coroot pairings. This describes representations of the Lie algebra, or of its simply connected Spin group; half-integral weights are allowed. For representations descending to $\mathrm{SO}(5)$ itself, the weight must lie in ${\mathbb{Z}}^2$, equivalently $b$ must be even.

The displayed torus-only Weyl integration formula needs $\phi$ to be a class function, meaning $\phi (xg{x}^{-1})=\phi (g)$. This hypothesis is implicit in the usual formula but is not printed here. We prove the general version for a continuous function, then specialize:Here $dg$ and $dt$ are probability Haar measures; $d\mu$ is the invariant probability measure on the homogeneous space $G/T$; and $W=N_G(T)/T$ is the finite Weyl group.

We use the following standard maximal-torus facts, as permitted: every element of a connected compact Lie group lies in some maximal torus; all maximal tori are conjugate; two elements of a fixed $T$ are conjugate in $G$ exactly when they are in the same $W$-orbit; and the complexified Lie algebra decomposes into ${\mathfrak{t}}_{\mathbb{C}}$ and one-dimensional root spaces. Opposite roots of a root system give real two-dimensional planes in ${\mathfrak{t}}^{\perp }$. At an element $t$ for which no root character has value one, the identity component of its centralizer is $T$.

Choose a positive root system ${\Phi }^{+}$. Regard each root $\alpha$ as a character $\alpha :T\to U(1)$ through the Adjoint representation. A convenient globally defined denominator isThe usual Weyl denominator additionally has a unit-modulus factor $e^{\rho }(t)$, with $\rho$ half the sum of positive roots. Where that half-weight is not a global torus character, it may be interpreted on a covering torus; the squared modulus is nevertheless unambiguous. In either convention,

Average an inner product on $\mathfrak{g}$ over the Adjoint representation to make it invariant. It defines a bi-invariant Riemannian metric on $G$ and the quotient metric on $G/T$. Consider the conjugation mapAt $(T,t)$, identify the source tangent directions with $X\in {\mathfrak{t}}^{\perp }$ and $Y\in \mathfrak{t}$, and translate the target tangent vector to the identity by $t^{-1}$. Differentiating givesThe two summands are orthogonal. On a real root plane, $\mathrm{Ad}(t^{-1})$ is a rotation with complex eigenvalues $\alpha (t{)}^{-1},\alpha (t)$. Thus the absolute real Jacobian determinant on this plane is $|1-\alpha (t){|}^2$. The torus direction contributes one. Therefore the conjugation Jacobian for a compact Lie group isThis is also the Jacobian for the normalized measures: the Riemannian submersion has fibers isometric to $T$, so $\mathrm{vol}(G)=\mathrm{vol}(G/T)\mathrm{vol}(T)$; the probability-volume normalization factors cancel.

Let $T_{\mathrm{reg}}$ be the elements for which $J_q\ne 0$. The restriction of $q$ is a local diffeomorphism onto the regular elements of $G$. It is $|W|$-to-one. Indeed, fixing $t_0\in T_{\mathrm{reg}}$, any pair with $xt{x}^{-1}=t_0$ has $xT{x}^{-1}=T$, because conjugation identifies the identity components of the two centralizers. Thus $x\in N_G(T)$, and there is exactly one pair $(xT,x^{-1}{t}_{0}x)$ for each coset in $N_G(T)/T$. This argument does not require the full centralizer of $t_0$ to be connected.

The nonregular locus has measure zero. In $T$ it is a finite union of sets $\alpha (t)=1$ of smaller dimension; their conjugation images are images of smaller-dimensional manifolds under $q$, and hence have zero volume in $G$. The change-of-variables formula for the regular covering therefore givesFor a class function, the inner integral is $\phi (t)$, proving the required formula with its correct hypothesis.

Without conjugation invariance, the torus-only expression can be false. For $G=SU(2)$ with diagonal $T$, take $\phi (g)=|g_{12}{|}^2$. It vanishes on $T$, whereas its group integral is $1/2$: the two squared coordinates of a Haar-distributed unit column have equal means and sum to one. Thus the conjugacy average cannot simply be dropped.

For a Cartan-Weyl basis, the Jacobi identity givesThus the bracket has joint Cartan weight $\alpha +\beta$. If $\beta =-\alpha$, it belongs to the zero-weight space, which is the Cartan subalgebra, and so is a linear combination ${\tilde{\alpha }}_i{H}_i$. If $\alpha +\beta$ is a nonzero root, its root space is one-dimensional and the bracket is proportional to $E_{\alpha +\beta }$. If that nonzero sum is not a root, the bracket vanishes. This proves the asserted forms without assuming every pair of roots has a root sum.

With the stated normalization, put $H_{\alpha }=2{\alpha }_i{H}_i/{\alpha }^2$. The commutators becomeThe adjoint relations make $J_3=H_{\alpha }/2$, $J_{+}=E_{\alpha }$, $J_{-}=E_{-\alpha }$ the usual SU(2) Lie algebra generators. Hence every finite-dimensional SU(2) irreducible component has $J_3$ eigenvalues $m=-j,-j+1,\dots ,j$, with $2j$ an integer. Therefore $H_{\alpha }$ has integer eigenvalues $2m$, including in the finite-dimensional Adjoint representation. Algebraically this is the sl2 subalgebra associated with a root.

If $[H_{\alpha },X_{\lambda }]=\lambda X_{\lambda }$, another use of the Jacobi identity givesThe bracket is therefore zero or lies in the eigenvalue-$\lambda +2$ space. Within one irreducible SU(2) ladder it is proportional to the next chosen vector $X_{\lambda +2}$. If the full eigenspace has multiplicity, the weight statement is the general one; proportionality to one named vector presupposes the ladder choice. Similarly $E_{-\alpha }$ lowers this eigenvalue by two, and the ladder weights run from $-n$ to $n$ in steps of two.

For a root vector $E_{\beta }$, its $H_{\alpha }$ weight isThe adjoint SU(2) ladder therefore provesFor nonparallel $\alpha ,\beta$, raising and lowering $E_{\beta }$ move its joint Cartan weight through the root string $\beta +k\alpha$. The finite irreducible ladder has weights symmetric about zero. Starting at $m_{\alpha \beta }$, reaching its partner $-m_{\alpha \beta }$ requires $k=-m_{\alpha \beta }$. The corresponding nonzero vector lies in the root space of $\beta -m_{\alpha \beta }\alpha$, proving the Weyl reflection property:There is no zero joint root in this string when $\alpha ,\beta$ are nonparallel. Its one-dimensional root spaces prevent two irreducible ladders with the same weight parity from overlapping, so the ladder argument applies to the entire string. For $\beta =\pm \alpha$, reflection simply interchanges the two opposite roots, which already exist by the adjoint relation.

For angle and length restrictions, set $p=2\alpha \cdot \beta /{\alpha }^2$ and $q=2\alpha \cdot \beta /{\beta }^2$. These Cartan integers have the same sign when nonzero and obeyFor nonparallel, nonorthogonal roots, $pq$ is an integer strictly between zero and four. The possibilities are consequently:

Orthogonal roots have $p=q=0$ and $\theta =\pi /2$. Parallel roots have $pq=4$; the integer pairs would allow only length multiples $1,2,1/2$. The stated reducedness condition excludes the latter two, so parallel roots are equal or opposite, with angles zero or $\pi$ and equal lengths. ThusNot every root system realizes every listed angle.

The orthogonal pair alone gives no ratio through $q/p$. To finish the length restriction for a simple Lie algebra, use irreducibility of its root system. Let $U$ span the Weyl group orbit of $\alpha$. Both $U$ and $U^{\perp }$ are invariant under the orthogonal root reflections. If a root had nonzero projections in both, reflection of its projection in $U$ would have a nonzero $U^{\perp }$ component, violating invariance. Hence every root is entirely in one of these subspaces. A proper nonzero $U$ would split the root system into two orthogonal parts, contradicting irreducibility. This proves that the Weyl orbit spans an irreducible root space.

In particular, some orbit root $\gamma$ is nonorthogonal to any chosen $\beta$, and ${\gamma }^2={\alpha }^2$. Applying the previous integer-pair restriction to $\gamma ,\beta$ therefore also handles orthogonal $\alpha ,\beta$:There can be at most two different root lengths: three lengths would have ratios two and three relative to the shortest, but then the longest-to-middle squared-length ratio would be $3/2$, which is forbidden. A reducible collection of orthogonal rank-one factors could have independently chosen scales; the simple-algebra assumption excludes that exception.

Choose a generic linear functional to define positive roots. The simple roots are those positive roots which cannot be written as sums of two positive roots; they form a basis in which all root coefficients are integers of one sign. Distinct simple roots have nonpositive inner product: if their inner product were positive, their root string would contain their difference, and whichever difference is positive would decompose one of them. In rank two, irreducibility excludes the orthogonal case. The allowed simple-root angles are therefore $120^{\circ },135^{\circ },150^{\circ }$, giving the three irreducible diagrams in the classification of rank-two root systems: A2 root system, B2 root system and G2 root system. Reflections in the two simple roots generate their respective six, eight and twelve roots.

Here are explicit simple-root choices and positive roots; adding their negatives gives each complete diagram.

The nonnegative coefficients in each positive-root list verify the chosen simple roots explicitly. The orthogonal four-root system $A_1\times A_1$ is rank two but reducible, so it is not a diagram for the simple algebra assumed here.

Define the Cartan matrix in the column-coroot convention:Using the ordered roots above givesThe alternative row-coroot convention transposes these matrices. Stating the convention and root ordering avoids a spurious disagreement in the unequal-length cases.

For the chosen simple roots $\Delta$, the open fundamental chamber of a root system isEquivalently $\lambda (h_{\alpha })>0$ for every simple coroot. Since all positive roots are nonnegative combinations of simple roots, requiring the inequality for all positive roots gives the same chamber. It is a connected component of the complement of the root reflection hyperplanes; the Weyl group permutes those chambers simply transitively.

The Closed dominant Weyl chamber is obtained by replacing the inequalities by $\ge 0$. It is the cone generated by the fundamental weights. When the fundamental Weyl chamber is taken to include its walls, this closed convention is intended; both conventions determine the same positive system and boundary rays.

The drawn simple roots have $\Vert \alpha {\Vert }^2=1$, $\Vert \beta {\Vert }^2=2$ and $\alpha \cdot \beta =-1$. Their Cartan integers are therefore $⟨\beta ,{\alpha }^{\vee }⟩=-2$ and $⟨\alpha ,{\beta }^{\vee }⟩=-1$, identifying a B2 root system. The corresponding root reflections act on coordinates byIn particular $s_{\beta }(\alpha )=\alpha +\beta =(0,1)$ and $s_{\alpha }(\beta )=2\alpha +\beta =(1,1)$, so both new positive roots are forced. Reflection in any root also forces its negative. ThusThese eight roots are drawn below, with the given simple pair highlighted.

For completeness, no further roots can occur. In a finite reduced root system, every root is a Weyl group translate of a simple root: reflect a nonsimple positive root in a simple root with positive inner product to decrease its height, and iterate. This standard simple-root descent may also be quoted here. The eight-vector set just listed is stable under both generating reflections, and the orbits of the two simple roots already fill it. Hence no simple-root orbit, and therefore no root, lies outside that set. This establishes both the necessity of every drawn root and exhaustion of the root system.

Let $\mathfrak{g}$ be a finite-dimensional complex semisimple Lie algebra, choose a Cartan subalgebra $\mathfrak{t}$ and a set $R^{+}$ of positive roots, and write $W$ for the Weyl group. Let $\lambda$ be a dominant integral weight and $L(\lambda )$ the finite-dimensional irreducible highest-weight representation with that highest weight. If $L(\lambda {)}_{\mu }$ denotes its weight space of weight $\mu$, its formal character is $\mathrm{ch}L(\lambda )={\sum }_{\mu }\dim L(\lambda {)}_{\mu }{e}^{\mu }$. The symbols $e^{\mu }$ belong to the group algebra of the weight lattice and multiply by $e^{\mu }{e}^{\nu }=e^{\mu +\nu }$; they are formal symbols, not matrix exponentials.

Put $\rho =\frac{1}{2}{\sum }_{\alpha \in R^{+}}\alpha$, the half-sum of positive roots, and let $\ell (w)$ be the Coxeter length of $w$, the least number of simple reflections expressing it. Then $(-1{)}^{\ell (w)}=\det (w)$ in the real reflection representation. The Weyl character formula statesBy the Weyl denominator formula, the denominator is $e^{\rho }{\prod }_{\alpha \in R^{+}}(1-e^{-\alpha })$, giving the equivalent expressionThe quotient can be interpreted in the fraction field of the formal group algebra; the theorem says it simplifies to the finite formal character above. In particular the apparent denominator does not mean that the representation has infinitely many weights.

The classification of finite-dimensional sl2 representations identifies $L(n)$ with ${\mathrm{Sym}}^n{\mathbb{C}}^2$. In the defining two-dimensional representation,so $e^2=f^2=0$ andOn the $n$th tensor power, the infinitesimal operator associated with $e$ is the sum of its actions on the individual factors. These actions commute, so exponentiating their sum is the tensor product of the individual exponentials; the same holds for $f$. Restricting to the symmetric power consequently identifies the operator $s$ on $L(n)$ with ${\mathrm{Sym}}^{n}S$. It follows thatIndeed $-I$ contributes one minus sign from each of the $n$ tensor factors. The Weyl reflection lift in an sl2 representation can therefore square to minus the identity, even though its image in the Weyl group is an involution.

The Weyl group acts on the weight lattice by its root reflections and extends to ring automorphisms of the group ring of a weight lattice:Weights and their multiplicities in a finite-dimensional Lie algebra representation of a semisimple Lie algebra are permuted by this action, so its formal character is Weyl-invariant.

Choose a set ${\Phi }^{+}$ of positive roots and putHere ${\omega }_i$ are the fundamental weights, and $\det (w)=(-1{)}^{\ell (w)}$, where the Coxeter length $\ell (w)$ counts positive roots carried to negative ones. For a dominant integral weight $\lambda$, let $L(\lambda )$ denote the finite-dimensional Irreducible Lie algebra representation of highest weight $\lambda$. The Weyl character formula isThe second denominator expression is the Weyl denominator formula. Both alternants change sign under each root reflection; their quotient is invariant. Although written as a fraction, the theorem asserts that it is a finite element of $\mathbb{Z}[{\Lambda }_W]$ with the nonnegative coefficients given by weight multiplicities. The highest-weight classification specifies which $\lambda$ label the irreducibles, while the root system and choice of positive roots determine every other term.

Work with a finite-dimensional complex semisimple Lie algebra. The Dynkin diagram is obtained from its intrinsic root system, with several structure theorems entering the construction.

First choose a Cartan subalgebra $\mathfrak{h}$. The semisimple structure theory gives the root-space decompositionThe nonzero roots span ${\mathfrak{h}}^{\ast }$, root spaces have dimension one, and every root has its negative. The rank is $r=\dim \mathfrak{h}$.

Next use the Killing form. Its nondegeneracy on $\mathfrak{h}$ and its positive-definite restriction to the real coroot span identify the real root span $E=\mathbb{R}\Phi$ with a Euclidean space. Explicitly, if $t_{\lambda }={\mathcal{B}}^{-1}(\lambda )$, put $(\lambda ,\mu )=B(t_{\lambda },t_{\mu })$. The sl2 subalgebra associated with a root supplies its normalized coroot $H_{\alpha }$ and the integral pairingThe root-string theorem makes the reflection $s_{\alpha }(\beta )=\beta -\beta (H_{\alpha })\alpha$ permute $\Phi$. Together with the absence of multiples other than $\pm \alpha$, this shows that $\Phi$ is a finite reduced root system and a crystallographic root system. These properties come from semisimple Lie algebra theory, not from the diagram's definition.

Choose a vector in $E$ that is not perpendicular to any root. Define the positive roots by positive inner product with this vector. The simple roots are those positive roots that cannot be expressed as sums of two positive roots. Root-system theory shows that they form a basis $\Delta =\{{\alpha }_1,\dots ,{\alpha }_r\}$, and each root is an integer combination of them with coefficients all of one sign. Distinct simple roots have nonpositive inner product.

Use the standard Cartan matrix conventionThere is one diagram vertex for each simple root. For $i\ne j$, the integral off-diagonal entries are nonpositive andConnect the two vertices with respectively zero, one, two or three edges. The corresponding angles are $90^{\circ },120^{\circ },135^{\circ },150^{\circ }$. A multiple edge carries an arrow toward the shorter root. For joined vertices, the length ratio is determined by $A_{ij}/A_{ji}=({\alpha }_i,{\alpha }_i)/({\alpha }_j,{\alpha }_j)$; hence the diagram records both angle and relative length.

For example, ${\mathfrak{sl}}_3$ has simple roots $L_1-L_2$ and $L_2-L_3$, giving two equal-length vertices joined by one edge: type $A_2$. For ${\mathfrak{sp}}_4$, take ${\alpha }_1=L_1-L_2$ and ${\alpha }_2=2L_2$. Thenso the two vertices have a double edge directed toward ${\alpha }_1$, the shorter root: type $C_2$.

Finally, orthogonal irreducible components of the root system correspond to the simple ideals of a Lie algebra in the semisimple Lie algebra, so its diagram is the disjoint union of their connected diagrams. Conjugacy of Cartan subalgebras and the Weyl group action on choices of positive roots make the resulting diagram independent of these choices up to isomorphism. Thus the vertices, edge multiplicities, arrows and connected components encode the simple-root geometry and the simple-ideal decomposition.

Choose a fundamental system of a root system $\{\alpha ,\beta \}$. Distinct simple roots have nonpositive inner product. Their Cartan matrix has diagonal entries two and off-diagonal entries $-p,-q$, where either $p=q=0$, or $p,q$ are positive integers. The root-system finiteness lemma gives $pq<4$. After exchanging the simple roots, the possibilities areTheir angles and relative lengths determine the two simple roots up to an orthogonal transformation and common scaling, except that the two orthogonal components can be scaled separately.

To see that no extra roots remain unspecified, write a positive root $\gamma =m\alpha +n\beta$ with nonnegative integer coefficients. If it is not simple, some simple root $\delta$ has $(\gamma ,\delta )>0$: otherwise $\Vert \gamma {\Vert }^2=m(\gamma ,\alpha )+n(\gamma ,\beta )\le 0$. The root reflection in $\delta$ subtracts a positive integer multiple of $\delta$. The other coordinate stays nonnegative; since $\gamma$ is not proportional to $\delta$, it is positive, so the reflected root must still be positive by the defining sign property of a fundamental system of a root system. Its height decreases. Repeating reduces $\gamma$ to a simple root. Thus every root is in a Weyl group orbit of a simple root.

For the four matrices, applying their simple reflections gives, respectively, the following positive roots:where in the last two rows $\alpha$ is long, $\beta$ short. For example, $s_{\beta }(\alpha )=\alpha +q\beta$ and $s_{\alpha }(\beta )=\beta +\alpha$ for the displayed matrices when the convention is $a_{ij}=2({\alpha }_i,{\alpha }_j)/({\alpha }_i,{\alpha }_i)$. The sets comprising these positive roots and their negatives are invariant under both reflections, as direct substitution verifies. The height argument therefore shows they are the whole root system. This proves the classification of rank-two root systems:

For a root $\alpha$, choose normalized root vectors $e_{\alpha }\in {\mathfrak{g}}_{\alpha }$, $f_{\alpha }\in {\mathfrak{g}}_{-\alpha }$ and the coroot element $h_{\alpha }\in \mathfrak{h}$ withThe requested sl2 subalgebra associated with a root isThese bracket relations identify its basis with the standard sl2 triple. The notation here denotes this subalgebra, not the root reflection $s_{\alpha }$ used for the Weyl group.

The weight diagram of the Adjoint representation of ${\mathfrak{sl}}_3$ has its six roots $L_i-L_j$, $i\ne j$, each of multiplicity one, and zero of multiplicity two, from the Cartan subalgebra.

Restrict to the root subalgebra generated by $X=E_{12}$, $Y=E_{21}$ and $H=H_{12}$. The explicit decomposition into irreducible modules isThe first summand has $H$ weights $2,0,-2$. The second has weights $1,-1$, since $[Y,E_{13}]=E_{23}$. The third likewise has weights $1,-1$, since $[Y,E_{32}]=-E_{31}$. The last commutes with all three generators and is trivial. These subspaces are stable under the sl2 triple, are independent and have dimensions totaling eight, so they exhaust the Adjoint representation.

The Killing form value is the trace of $\mathrm{ad}{E}_{12}\mathrm{ad}{E}_{21}$. By the formula just proved, the four summands contribute $4,1,1,0$. Therefore

Identify the real span of the roots with its dual using the invariant inner product, and put ${\alpha }^{\vee }=2\alpha /(\alpha ,\alpha )$. The weight lattice isThe Weyl group is generated by the root reflections $s_{\alpha }$. Each reflection permutes the finite set of roots and preserves their lengths, hence also permutes their coroots. Thus, for $\lambda \in {\Lambda }_W$,Every generating reflection therefore maps the weight lattice to itself, and so does every product. Since the inverse product has the same property, every Weyl-group element acts bijectively on ${\Lambda }_W$. The roots span the ambient space, so the action on the finite root set is faithful; in particular the Weyl group is finite.

Choose simple roots ${\alpha }_1,\dots ,{\alpha }_r$. For the orbit statement one must use the Closed dominant Weyl chamberIts interior, defined using strict inequalities, is the open Weyl chamber. Weights on reflecting walls, including zero, cannot be moved into that interior, so closure is necessary here.

Choose $\rho$ strictly inside the chamber, for example the sum of the fundamental weights. The finite Weyl group orbit of $\beta$ contains an element $\lambda$ maximizing $(\lambda ,\rho )$. If $(\lambda ,{\alpha }_i)<0$ for some simple root, thencontradicting the maximum. Thus every simple-root pairing of $\lambda$ is nonnegative, and $\lambda \in \mathcal{W}$. Consequently some $w\in W$ sends every integral weight $\beta$ into the closed fundamental chamber. The argument in fact works for every vector in the real weight space.

For $G=U(n)$ take the diagonal maximal torusThe spectral theorem for unitary matrices says that every element is conjugate to an element of $T$. A diagonal element with pairwise distinct eigenvalues has centralizer exactly $T$. Its conjugates therefore determine its eigenvalues up to permutation, and the normalizer quotient is $W=N_G(T)/T\cong S_n$. The character lattice of a torus $T$ is ${\mathbb{Z}}^n$, with $e^{\lambda }(z)=z_1^{{\lambda }_1}\cdots z_n^{{\lambda }_n}$. Conjugation on the matrix unit $E_{ij}$ has weight $e_i-e_j$, so these are the roots. Choose $e_i-e_j$ positive for $i<j$; thenDominant integral weights are precisely the integer tuples ${\lambda }_1\ge \cdots \ge {\lambda }_n$. Negative entries are allowed: they correspond to determinant twists, so polynomial representations alone do not exhaust the irreducible representations of $U(n)$.

Normalize Haar measure on $G$ and $T$ to have mass one, and putThe Weyl integration formula for U(n) for a continuous class function isFor a general continuous function, replace $f(t)$ in the torus integral by ${\int }_{G/T}f(gt{g}^{-1})d(gT)$, using the normalized invariant quotient measure.

To prove the formula, consider the conjugation map $q:G/T\times T\to G$, $q(gT,t)=gt{g}^{-1}$. On regular torus elements its fibres have exactly $n!$ points: the possible orderings of the distinct eigenvalues. Equip the group with an invariant inner product on its Lie algebra and split $\mathfrak{u}(n)=\mathfrak{t}\oplus {\mathfrak{t}}^{\perp }$. After translating the derivative back to the identity, its torus directions contribute the identity, while its transverse directions contribute ${\mathrm{Ad}}_{t^{-1}}-I$. Each pair $i<j$ supplies a real two-dimensional off-diagonal plane. On that plane ${\mathrm{Ad}}_t$ is multiplication by $z_i/z_j$, regarded as a plane rotation, so the absolute real determinant of the transverse derivative isThe change-of-variables theorem, divided by the covering multiplicity, now gives the desired weighted integral up to the constant relating the normalized invariant measures. The nonregular locus has measure zero: repeated eigenvalues are the zero-set of the nonzero discriminant, and the torus coincidence sets likewise have measure zero. Thus its omission causes no change to the integral. To determine the normalization, expand $\Delta$ as a sum of $n!$ distinct torus characters. Fourier orthogonality gives ${\int }_T|\Delta {|}^{2}dt=n!$. Applying the formula to $f=1$ fixes the coefficient to $1/n!$. This proves both the class-function and conjugacy-averaged forms.

We can now derive the character formula from integration, rather than merely quote it. Use the basic highest-weight theorem: a finite-dimensional irreducible representation has a unique highest weight $\lambda$ with a one-dimensional highest-weight space; all its other weights are $\lambda$ minus a nonnegative sum of positive roots. Its character is Weyl invariant. Also, character orthogonality for compact groups gives ${\int }_G|{\chi }_{\lambda }{|}^2=1$. This orthogonality follows by averaging the representation on $\mathrm{End}(V_{\lambda })$: the integral projects onto intertwining operators, whose dimension is one by the Schur lemma.

Let $\delta =(n-1,n-2,\dots ,0)$ and, for a strictly decreasing integer tuple $\eta$, put $A_{\eta }(z)=\det (z_i^{{\eta }_j})$. The finite Laurent polynomial $\Delta {\chi }_{\lambda }$ is alternating. Every alternating Laurent polynomial is a linear combination of the $A_{\eta }$: collect its monomials by permutation orbits, and observe that an orbit with repeated exponents has zero coefficient by antisymmetry. Different $A_{\eta }$ are orthogonal on $T$ and each has squared norm $n!$.

The coefficient of $z^{\lambda +\delta }$ in $\Delta {\chi }_{\lambda }$ is exactly one. Indeed, a contribution from the term $z^{w\delta }$ would require a weight $\lambda +\delta -w\delta$ in $V_{\lambda }$. For $w\ne 1$, the vector $\delta -w\delta$ is a nonzero nonnegative combination of the positive simple roots, so that weight lies above $\lambda$ and is impossible. The identity permutation supplies the highest-weight coefficient one. ConsequentlyApply the integration formula to $|{\chi }_{\lambda }{|}^2$. Torus orthogonality givesAll the other coefficients vanish, provingThe quotient extends across repeated eigenvalues because its product with the denominator is the already defined continuous character. Equivalently, alternant divisibility makes it a Laurent polynomial invariant under permutations; multiply by a determinant power if negative exponents must first be removed.

Finally, $\delta -\rho =\frac{n-1}{2}(1,\dots ,1)$ is fixed by the Weyl group, so its common factor cancels in the quotient. Since $A_{\rho }=e^{\rho }{\prod }_{\alpha >0}(1-e^{-\alpha })$, the determinant expression is exactlyUsing $\delta$ in the derivation avoids treating a possibly half-integral $\rho$ as an actual character of the torus. At the identity, taking the leading Vandermonde term of the two determinants gives the accompanying dimension formula

The anti-Hermitian operators represent the compact real form; its complexification also contains the Hermitian $H_i$ and the raising operators and lowering operators $E_{\pm }^r$, with $(E_{+}^r{)}^{†}=E_{-}^r$. Define the roots in the two-dimensional Cartan plane byEach has squared length one. For each $r=1,2,3$, setThe displayed commutators give $[J_3^{(r)},J_{\pm }^{(r)}]=\pm J_{\pm }^{(r)}$ and $[J_{+}^{(r)},J_{-}^{(r)}]=2J_3^{(r)}$. Thus these are the three root SU(2) subalgebras of SU(3).