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A principal sl2 subalgebra of a semisimple Lie algebra has semisimple generator $h_{\mathrm{pr}}=2{\rho }^{\vee }$. Restriction to it packages a representation's weights into ordinary ${\mathfrak{sl}}_2$ weight strings.

The infinitesimal embedding is a subalgebra of the semisimple Lie algebra; integrating it to a Lie-group homomorphism is a separate group-level construction.

The quadratic Casimir element is a central element of $U(\mathfrak{g})$. On a highest-weight module of highest weight $\lambda$, one standard normalization makes it act by $|\lambda +\rho {|}^2-|\rho {|}^2$.

A highest-weight representation is generated by a weight vector killed by every positive-root subspace.

The generating vector has a highest Weight relative to the chosen positive root system.

For a complex semisimple Lie algebra, a Cartan subalgebra is a maximal abelian subalgebra consisting of semisimple elements. It is also called a maximal torus in this setting.

A nonabelian Lie algebra is simple when its only ideals are zero and the entire algebra.

For a choice of positive roots, the corresponding Borel subalgebra is $\mathfrak{b}=\mathfrak{t}\oplus {\mathfrak{n}}^{+}$, the sum of a Cartan subalgebra and all positive root spaces.

A semimartingale has a decomposition $X=X_0+M+A$ into a local martingale $M$ and an adapted finite-variation process $A$. Under standard normalizations the decomposition is unique.

A process $A$ has finite variation on compact intervals when every sample path has finite total variation of a function there. Such a process can be integrated pathwise by the Lebesgue-Stieltjes integral.

For the even potentialthe turning points are $\pm 4E$ for $0\le E\le 1$ and $\pm (E+1{)}^2$ for $E\ge 1$. Its action is

The internal effective temperature parametrizes the intrinsic cooling luminosity of a planet through $F_{\mathrm{int}}=\sigma T_{\mathrm{int}}^4$.

The irradiation temperature is defined from the incident stellar flux, commonly by $T_{\mathrm{irr}}=T_{\ast }\sqrt{R_{\ast }/a}$ before redistribution and albedo factors are applied.

Two Seifert forms are algebraically concordant when their difference is metabolic. Stable equivalence classes form the algebraic concordance group.

A Seifert matrix represents the Seifert form in an integral basis of $H_1(F)$. Its skew-symmetrization $A-A^T$ represents the intersection form of the surface and is unimodular.

Let ${\pi }_k(x)$ count primes at most $x$ congruent to $k$ modulo eight. Bounding the prime-power divisibility ofabove and its size below gives, for all sufficiently large $x$,

If $q_1,\dots ,q_r$ listed all primes congruent to seven modulo eight, then for $n=8q_1\cdots q_r$ the odd number $(n^2-2)/2$ is seven modulo eight. Every prime factor is one or seven modulo eight, at least one is seven, and none is among the $q_i$, a contradiction.

For $z=z(x)$, the second derivative transforms as $w_{xx}=z_x^2{w}_{zz}+z_{xx}{w}_z$.

The equation $(1-x^2)y\prime \prime -2xy\prime +\ell (\ell +1)y=0$ has polynomial solutions for nonnegative integer $\ell$.

The equation $(1-x^2)y\prime \prime -2xy\prime +\ell (\ell +1)y=0$ has the Legendre polynomial of degree $\ell$ as its normalized regular polynomial solution when $\ell$ is a nonnegative integer.

Given one nonzero solution of a homogeneous second-order linear equation, reduction of order constructs a second by writing it as a variable multiple of the first.

Variation of parameters replaces the constants in a complementary solution by functions to construct a particular solution.

For the inhomogeneous linear differential equation $y^{\prime \prime }+p{y}^{\prime }+qy=r$, assume $p,q,r$ are continuous functions on an interval $I$, and $y_1,y_2$ are linearly independent homogeneous solutions there. With $W=W(y_1,y_2)$ and $x_0\in I$, variation of parameters constructs the particular solutionThe Wronskian has no zeros on $I$, so these quotients are defined. For $A{y}^{\prime \prime }+B{y}^{\prime }+Cy=F$, normalize first: $r=F/A$, not $F$.^[ref]

To derive the formula, write $y_p=u_1{y}_1+u_2{y}_2$ and impose $u_1^{\prime }{y}_1+u_2^{\prime }{y}_2=0$. This auxiliary choice removes extra terms from the first derivative. Substitution into the inhomogeneous linear differential equation leaves $u_1^{\prime }{y}_1^{\prime }+u_2^{\prime }{y}_2^{\prime }=r$. The two linear equations for $u_1^{\prime },u_2^{\prime }$ have determinant $W$, and Cramer's rule givesIntegrating from $x_0$ makes $y_p(x_0)=y_p^{\prime }(x_0)=0$. Arbitrary integration constants contribute only homogeneous solutions; adding $c_1{y}_1+c_2{y}_2$ supplies the general solution.

For a worked example, use the same homogeneous solutions $e^{-x},e^{2x}$ and the forcing term $r=6e^{3x}$ in $y^{\prime \prime }-y^{\prime }-2y=6e^{3x}$. With $x_0=0$ and $W=3e^x$,Consequently the particular solution selected by zero initial conditions isIts value and first derivative vanish at zero. Direct substitution confirms the forcing term: the differential operator $L[y]=y^{\prime \prime }-y^{\prime }-2y$ annihilates the last two terms and maps $e^{3x}$ to $4e^{3x}$. The two integrals therefore build a response from known homogeneous solutions, even when the forcing term has no convenient elementary guess.

For two differentiable functions $f,g$ on an interval $I$, their Wronskian is the functionThe order matters: exchanging $f$ and $g$ reverses the sign. For $n$ sufficiently differentiable functions, the Wronskian is the determinant $\det (f_j^{(i-1)}(x){)}_{i,j=1}^n$.^[ref]

The Wronskian measures whether the vectors $(f(x),f^{\prime }(x))$ and $(g(x),g^{\prime }(x))$ supply independent value-and-derivative data at that point. For real-valued functions, its absolute value is the area of the parallelogram they span. If a constant linear combination $af+bg$ is zero throughout $I$, then $a{f}^{\prime }+b{g}^{\prime }=0$ too. A nonzero Wronskian at even one point makes this pair of linear equations invertible and forces $a=b=0$. Thus a nonzero Wronskian somewhere always proves linear independence of the functions on $I$.

For an explicit example, the exponential functionssolve the same homogeneous linear differential equation $y^{\prime \prime }-y^{\prime }-2y=0$ on the whole real line: substitution gives $1+1-2=0$ and $4-2-2=0$, respectively. Their Wronskian isFor the initial value problem $y(0)=2$, $y^{\prime }(0)=-1$, the linear combination $y=c_1{y}_1+c_2{y}_2$ requires $c_1+c_2=2$ and $-c_1+2c_2=-1$. Hence $c_1=5/3$, $c_2=1/3$, and $y=\frac{5}{3}{e}^{-x}+\frac{1}{3}{e}^{2x}$. The nonzero Wronskian makes the two initial conditions independently adjustable.

The stronger converse requires a common equation. Suppose $y_1,y_2$ solve $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0$ with continuous functions $p,q$ on one interval $I$. Then the functions are linearly independent if and only if their Wronskian is nonzero at one point, equivalently at every point. If $W(x_0)=0$, choose constants $a,b$, not both zero, making $z=a{y}_1+b{y}_2$ satisfy $z(x_0)=z^{\prime }(x_0)=0$. The superposition principle makes $z$ a solution of a differential equation for the same equation, and the uniqueness theorem for ordinary differential equations forces $z$ to be identically zero. This proves linear dependence.^[ref]

For arbitrary differentiable functions, an identically zero Wronskian need not imply linear dependence. Consider, on $I=(-2,2)$,Both functions have three continuous derivatives. On each side of zero, $g$ is a constant multiple of $f$, so $W(f,g)=0$; at zero their values and first derivatives are all zero, so the Wronskian is zero there too. Nevertheless $af+bg=0$ evaluated at $x=-1$ gives $a-b=0$, while evaluation at $x=1$ gives $a+b=0$. Hence $a=b=0$: these functions are linearly independent on the whole interval. There is no single constant multiplier valid on both sides of the common zero.

These functions cannot both solve any equation $y^{\prime \prime }+p{y}^{\prime }+qy=0$ with continuous functions $p,q$ throughout $(-2,2)$: even $f$ alone would contradict uniqueness because $f(0)=f^{\prime }(0)=0$ but $f$ is not identically zero. They do solve the singular equation $x^2{y}^{\prime \prime }-3x{y}^{\prime }=0$, whose leading coefficient vanishes at zero. Dividing by $x^2$ gives $p=-3/x$ only away from zero; Abel's identity cannot be applied across that point. The example pinpoints why checking the equation and its interval is essential before using a zero Wronskian to infer linear dependence.