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For the defining representation of the Special orthogonal Lie algebra, Hermitian Lie algebra generators $t_{ij}=i(E_{ij}-E_{ji})/\sqrt{2}$ have trace normalization $\mathrm{tr}(t_{ij}{t}_{kl})={\delta }_{ik}{\delta }_{jl}$ for ordered pairs. Summing their squares gives ${\sum }_{i<j}{t}_{ij}^2=(N-1)I/2$. Consequently the Casimir eigenvalue is $(N-1)/2$ and the Dynkin index is one. Doubling the trace normalization doubles both numbers; a gauge coupling must be rescaled inversely with the Lie algebra generators. With long roots of squared length two, the first convention is $C_2(\lambda )=(\lambda ,\lambda +2\rho )/2$.

The pressure of fluid in pore space. On fluid-accessible contacts, it reduces effective compression and can promote frictional sliding or tensile opening at fixed total stress.

For an on-shell W with zero transverse recoil and massless leptons, $p_T=(M_W/2)\sin \theta$. Mapping both angular roots to one transverse momentum produces an integrable inverse-square-root edge. For a normalized polar distribution $3(1-\cos \theta {)}^2/8$, the density in $x=2p_T/M_W$ is $3x(2-x^2)/(4\sqrt{1-x^2})$. Finite width, transverse recoil and detector resolution smear the edge.

A diagram of stable mineral assemblages as functions of specified aqueous thermodynamic activities, at fixed temperature and pressure. The coordinates often use logarithmic activity ratios. Mass-action equations give pairwise coexistence lines; a line segment is stable only when no omitted mineral has a lower relevant thermodynamic potential under the same imposed chemical potentials.

An ultramafic rock dominated by olivine and orthopyroxene; strongly melt-depleted mantle commonly has this composition.

Suppose both $(a_n)$ and $(-a_n)$ have only finitely many peaks of a sequence. Beyond some index $N$, every term has a strictly larger later term and a strictly smaller later term. Starting from $a_N$, construct one strictly increasing and one strictly decreasing monotone subsequence.

If the original sequence had limit of a sequence $L$, both selected subsequences would tend to $L$. The increasing one has a second term strictly above $a_N$, so $L>a_N$; the decreasing one similarly forces $L<a_N$. This contradiction proves nonconvergence. For example, $a_n=(-1{)}^{n}n$ has no peaks of a sequence in either sign and diverges.

Suppose $k_1<k_2<\cdots$ are peaks of a sequence $(a_n)$ and $m_1<m_2<\cdots$ are peaks of a sequence $(-a_n)$. Then $(a_n)$ converges precisely when these two subsequences have a common finite limit of a sequence $L$.

Necessity follows because every subsequence inherits the limit of a sequence of a convergent sequence. For sufficiency, choose one selected value from each subsequence within $\epsilon$ of $L$. Every term after both selected indices lies between the two selected values:This is exactly $a_n\to L$.