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For a homogeneous harmonic polynomial $H_l$ of degree $l$ in three dimensions, the Euler theorem for homogeneous functions gives $D{H}_l=l{H}_l$, where $D=\mathbf{x}\cdot \nabla$. The orbital angular momentum identity ${\hat{L}}^2=-{\hslash }^2(r^2\Delta -D^2-D)$ therefore gives ${\hat{L}}^2{H}_l=l(l+1){\hslash }^2{H}_l$. This constructs angular momentum eigenstates directly from polynomials.

For a single-component equilibrium system with extensive internal energy $E(S,V,N)$, scaling its entropy, volume and particle number together gives $E(\lambda S,\lambda V,\lambda N)=\lambda E(S,V,N)$. The Euler theorem for homogeneous functions therefore gives $E=S{\partial }_{S}E+V{\partial }_{V}E+N{\partial }_{N}E$. The first law of thermodynamics identifies these derivatives as $T,-P,\mu$, respectively, proving $E=TS-PV+\mu N$. Dividing by volume yields the entropy density formula $s=(\rho +P-\mu n)/T$. This step fixes the entropy normalization rather than merely identifying its differential. Differentiating the Euler relation and subtracting $dE=TdS-PdV+\mu dN$ gives the Gibbs-Duhem equation $SdT-VdP+Nd\mu =0$. The argument requires extensivity; it is not an assumption about arbitrary nonadditive systems.

The Euler theorem for homogeneous functions applied to an extensive quantity, together with $dE=TdS-pdV+\mu dN$ give

A function is homogeneous of degree $k$ when rescaling its arguments by a common scalar $\lambda$ rescales its value by ${\lambda }^k$, whenever the arguments and that power are defined. For integer $k$ over a vector space, one can require every nonzero scalar; a homogeneous polynomial gives an example. On a domain stable under positive real rescaling, the positive-scaling version allows real $k$. Degree-one positive homogeneity is the convention used for norms and many convex functions. The Euler theorem for homogeneous functions turns this scaling law into a differential identity for a differentiable function.

Let $K=-3\log \Gamma$ where $\Gamma >0$ is twice differentiable and homogeneous of degree one. If the Kähler Hessian is invertible, the Euler theorem for homogeneous functions identities give ${\tau }_i{K}_{ij}=3{\Gamma }_j/\Gamma$, then $K_{ij}^{-1}{\Gamma }_j/\Gamma ={\tau }_i/3$ and ${\Gamma }_i{K}_{ij}^{-1}{\Gamma }_j/{\Gamma }^2=1/3$. Multiplication by nine gives the no-scale identity. A positive degree-one $\Gamma$ need not induce an invertible or positive Kähler metric: $\Gamma ={\tau }_1+{\tau }_2$ gives a rank-one metric for $K$. The Hessian matrix $({\Gamma }_{ij})$ itself always has the radial null vector $\tau$, whereas the Hessian matrix of $K$ can be invertible. In a larger theory the relevant full inverse metric, or a decoupled block, must obey the identity.

For a box $(i,j)$ of a Young diagram, its hook of a Young diagram contains that box, all boxes to its right in its row, and all boxes below it in its column. Its hook length is $h_{ij}={\lambda }_i-j+{\lambda }_j^{\prime }-i+1$. The hook graph of a partition is the diagram with each box labeled by its hook length. Write $H_{\lambda }={\prod }_{(i,j)\in \lambda }{h}_{ij}$ for the hook product of a partition. The hook-length formula is

Pad $\lambda$ to $m$ rows and use the beta set of a partition ${\ell }_i={\lambda }_i+m-i$, so ${\ell }_1>\cdots >{\ell }_m\ge 0$ and ${\sum }_i{\ell }_i=n+d$, where $d=\left(\genfrac{}{}{0ex}{}{m}{2}\right)$. We prove the beta-set hook-product identity row by row. For $1\le j\le {\lambda }_i$, put $q_j=m+j-{\lambda }_j^{\prime }-1$; then $h_{ij}={\ell }_i-q_j$. The $q_j$ are distinct integers in $[0,{\ell }_i-1]$, and none equals a beta number. Indeed, if $k\le {\lambda }_j^{\prime }$ then ${\ell }_k\ge q_j+1$, whereas if $k>{\lambda }_j^{\prime }$ then ${\ell }_k\le q_j-1$. Exactly $m-i$ beta numbers lie below ${\ell }_i$, so the remaining ${\lambda }_i$ integers in that interval are precisely these $q_j$. ThereforeThis gives the equivalent Specht module dimension expression $f^{\lambda }=n!\Delta (\ell )/{\prod }_i{\ell }_i!$.

The standard Young tableaux of shape $\lambda$ form the dimension count for the Specht module. The largest entry $n$ must occupy a removable corner. Deleting it bijects tableaux with the disjoint union of standard tableaux of the shapes obtained by deleting a corner, provingSet a term to zero whenever $\lambda -e_i$ is not a partition: this includes equal adjacent rows and an attempted deletion from a zero row. The empty diagram has dimension $1$.

For an algebraic proof that the proposed formula has the same recurrence, establish the Vandermonde shift identityIts left side is an alternating polynomial in the $x_i$, because permuting the variables permutes the summands and changes the sign of every Vandermonde factor. It is therefore divisible by $\Delta (x)$. The quotient is symmetric in $x$ and homogeneous of total degree one in $(x,t)$, so has the form $a{\sum }_i{x}_i+bt$. At $t=0$, $a=1$. Differentiating in $t$ at zero and using the Euler theorem for homogeneous functions gives ${\sum }_i{x}_i{\partial }_i\Delta =d\Delta$, so $b=d$. This proves the identity as a polynomial identity, including repeated coordinates.

Take $x=\ell$ and $t=-1$. Since ${\sum }_i{\ell }_i=n+d$, it gives ${\sum }_i{\ell }_i\Delta (\ell -e_i)=n\Delta (\ell )$. For $n\ge 1$, this is exactlyThe summand is the proposed dimension for $\lambda -e_i$; if two beta numbers collide its Vandermonde is zero, and if ${\ell }_i=0$ its coefficient is zero, so no negative factorial is needed. For the empty partition, $\ell =(m-1,\dots ,0)$ and $\Delta (\ell )={\prod }_{k=0}^{m-1}k!$, giving initial value $1$. Induction now proves the hook-length formula from the tableau recurrence.

For the final sum, tuples with repeated coordinates contribute zero. Sorting each distinct nonnegative tuple with sum $\left(\genfrac{}{}{0ex}{}{m+1}{2}\right)$ gives one beta set of a partition of size $m$, since subtracting the staircase removes $\left(\genfrac{}{}{0ex}{}{m}{2}\right)$ from the sum. Conversely every partition of $m$, padded to $m$ rows, supplies exactly $m!$ ordered tuples, with the same squared summand. Hence the square-sum identity for shifted partition coordinates isThe middle equality uses the Artin–Wedderburn theorem for the group algebra and the complete classification of its simple modules by Specht modules. It explains why the last identity is a representation-dimension count rather than an accidental cancellation.

The exponent in the original PDF is $e^{-\alpha S}$ with real $\alpha >0$, not the $e^{-aS}$ produced by the local TeX. Use that authoritative expression. PutThe domain conditions ensure a real Kähler potential and positive Kähler metric. There are no gauge multiplets specified, so the scalar potential is the supergravity F-term potentialThe Kähler covariant derivative of a superpotential and metric entries areState the inverse with its indices explicitly, since transposing the off-diagonal complex entries would change the answer:Here ${\sum }_j{K}_{i\bar{j}}{K}^{k\bar{j}}={\delta }_i^k$, so the displayed upper-index array is the transpose of the ordinary matrix inverse of the displayed lower-index array.

Substitution shows that the cross terms involving $W$ cancel and that the $T,C$ sector givesThe first term cancels the universal negative term. Thus the no-scale supergravity potential isThis positivity is an algebraic cancellation, not an assumption that the individual Kähler derivatives vanish.

Choose the supergravity auxiliary field convention $F^i=-e^{K/2}{K}^{i\bar{j}}\overline{D_{j}W}$. Direct multiplication givesA conventional common phase or overall sign on the auxiliary fields changes none of the vanishing conditions. A supersymmetric configuration requires every $F^i$ to vanish. First $F^C=0$ implies $C=0$, then $F^T=0$ implies $w=0$, and $F^S=0$ implies $w_S=0$. Since $w_S=-\alpha a{e}^{-\alpha S}$, at a finite point of the physical domain this is possible only when $a=0$, followed by $b=0$. Consequently If $(a,b)\ne (0,0)$, every finite configuration has a nonzero auxiliary field, so any finite vacuum breaks supersymmetry. The printed request cannot hold for arbitrary parameters without this exception: $a=b=0$, $C=0$ is a zero-energy supersymmetric family with both $S$ and $T$ unfixed.

With the stipulated vanishing vacuum expectation value of $C$, define $A(S)=b+(1+\alpha s)a{e}^{-\alpha S}$. The potential reduces to $|A{|}^2/(s{t}^3)$. Its derivative along $t$ is $-3V/t$, so a finite stationary point must have $A=0$. Such a point is a global minimum of the full nonnegative potential, since both squares vanish at $C=0$. The finite zero-energy vacuum for a single exponential superpotential therefore satisfiesFor nonzero $a,b$, write $S=\sigma +i\chi$ and $x=\alpha \sigma >0$. The modulus and phase conditions areThe function $(1+2x)e^{-x}$ increases up to $x=1/2$ and then decreases to zero; its maximum is $2e^{-1/2}$. Hence a finite minimum exists exactly whenThere is one positive solution for $0<\rho \le 1$, two for $1<\rho <2e^{-1/2}$, and one coalesced solution at the upper bound. The formal $x=0$ solution at $\rho =1$ is outside $s>0$. At each allowed $x$, the phase fixes $\chi$ modulo $2\pi /\alpha$. Equivalently $x=-\frac{1}{2}-W_k(-\rho /(2\sqrt{e}))$, using the real branches of the Lambert W function and retaining only $x>0$.

At any of these nontrivial minima, $w=-\alpha sa{e}^{-\alpha S}\ne 0$. The auxiliary fields becomeThis is a supersymmetry breaking Minkowski minimum in which the nonzero auxiliary field belongs to $T$. Both real components of $T$ are exact flat directions of a scalar potential at the minimum. They change the metric and auxiliary-field magnitudes but not the zero potential. Generically the two real components of $S$ are fixed. At the coalesced solution $x=1/2$, the radial quadratic restoring term vanishes, but the leading restoring term is quartic; this is not an additional exact flat direction of a scalar potential. The matter field $C$ likewise has a positive leading quartic potential, not an exact flat direction of a scalar potential, despite its zero quadratic mass here.

The exceptional parameter cases must also be stated. If $a=0,b\ne 0$ or $a\ne 0,b=0$, no finite zero-energy minimum with $C=0$ exists. The same holds when $|b/a|>2e^{-1/2}$. Because the $t$ derivative is nonzero at any positive-energy point on $C=0$, none is a finite minimum there; the energy approaches zero along the runaway $t\to \infty$. If $a=b=0$, all physical $S,T$ at $C=0$ give supersymmetric zero-energy minima. Thus an unconditional finite minimum or unconditional breaking would be a false claim for the printed arbitrary parameters.

For the homogeneous model, assume $\Gamma >0$ is twice differentiable, of degree one in the real moduli ${\tau }_i$, and that its Hessian matrix for $K$ is invertible on the sector considered. Differentiate $K=-3\log \Gamma$:The two Euler theorem for homogeneous functions identities in the question implyMultiplying by the inverse gives $K_{ij}^{-1}{\Gamma }_j/\Gamma ={\tau }_i/3$, so contracting once more yieldsThis is the no-scale identity from degree-one homogeneity. It does not assert that every degree-one function gives a positive or invertible metric: for example $\Gamma ={\tau }_1+{\tau }_2$ has a rank-one Kähler Hessian, and its inverse is undefined. The inverse hypothesis is necessary.

For complex moduli with ${\tau }_i=T_i+{\bar{T}}_i$, the same derivatives are the mixed Kähler metric; choosing real parts with a factor of two only introduces factors that cancel in the contraction. If $W$ is independent of these moduli, $D_{i}W=K_{i}W$, and their contribution is $3e^K|W{|}^2$. It cancels the universal $-3e^K|W{|}^2$, leaving no scalar potential from this isolated no-scale sector. Other chiral sectors can still contribute positive terms. For this conclusion in a larger theory, the displayed metric must be the appropriate decoupled no-scale block, or the full inverse metric must itself satisfy the corresponding identity; arbitrary mixed additions to $K$ do not inherit the cancellation automatically.

For a simple homogeneous thermodynamic system, internal energy is extensive: $E(\lambda S,\lambda V,\lambda N)=\lambda E(S,V,N)$. Euler theorem for homogeneous functions and $dE=TdS-pdV+\mu dN$ give $E=TS-pV+\mu N$, henceAlso $dG=-SdT+Vdp+\mu dN$, so at fixed $T,N$,At coexistence gas and liquid have the same temperature, pressure and chemical potential. Integrating $d\mu =vdp$ along the intervening isotherm gives $\int vdp=0$. Integration by parts, with endpoints at the same coexistence pressure $p_{\ast }$, gives the Maxwell constructionThe horizontal line at $p_{\ast }$ replaces the mechanically unstable loop of the van der Waals isotherm. The extensivity argument presupposes the ordinary thermodynamic limit and neglects macroscopic interface contributions.

Let $V$ be the physical volume of a region with fixed comoving coordinates, so $V\propto a^3$, and write $U=\rho V$. The first law of thermodynamics is $dU=TdS-PdV+\mu dN$. Use extensive quantities: under replication, $U(\lambda S,\lambda V,\lambda N)=\lambda U(S,V,N)$. The Euler theorem for homogeneous functions givesSetting $\mu =0$ yieldsThis is the entropy form of the Gibbs-Duhem equation; it fixes the normalization by extensivity, rather than adding an arbitrary volume-independent entropy constant.

For equilibrium expansion of the same region, the first law and the cosmological perfect-fluid continuity equation implyTherefore $\boxed{\dot{S}=0}$, establishing cosmological entropy conservation for adiabatic equilibrium evolution. Here “comoving volume” means a patch moving with the cosmological fluid; its physical volume $V$ changes. The corresponding coordinate volume is constant. The local TeX's $\dot{V}/P$ in the continuity-equation hint is a transcription error: the PDF correctly has $\dot{V}/V$.

The fundamental thermodynamic relation isBecause $E(S,V,N)$ is extensive, the Euler theorem for homogeneous functions givesSubstitution into the Gibbs free energy $G=E+pV-TS$ therefore yieldsFor a one-component equilibrium phase, the intensive chemical potential depends on the intensive state variables $T$ and $p$.

Differentiatewith respect to $\lambda$. The chain rule givesSetting $\lambda =1$ proves the Euler theorem for homogeneous functions

The left side is an alternating polynomial of total degree one more than $\Delta (x)$. Dividing by the Vandermonde determinant gives a symmetric homogeneous polynomial of degree one, necessarily $a{\sum }_i{x}_i+bt$. At $t=0$, $a=1$; differentiation in $t$ and the Euler theorem for homogeneous functions give $b=\left(\genfrac{}{}{0ex}{}{m}{2}\right)$. Thus the identity holds as a polynomial identity, even at repeated coordinates. At shifted partition coordinates with $t=-1$, it proves the removable-corner recurrence in the hook-length formula.