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The Baker--Campbell--Hausdorff formula expresses the local matrix logarithm of a product of exponentials as

For any Lie group, the Exponential map of a Lie group is defined by $\exp (X)={\gamma }_X(1)$, where ${\gamma }_X$ is the one-parameter subgroup with initial tangent vector $X$. For a Matrix Lie group, it is the matrix exponential $\exp X={\sum }_{n\ge 0}{X}^n/n!$. Its differential at zero is the identity, so the inverse function theorem supplies a local exponential chart. A small transformation can therefore be encoded by a vector in the Lie algebra. See MIT 18.745, Lecture 7, Definition 7.2 and Theorem 7.5.

Local invertibility makes no promise about a global inverse. In the planar rotation example, $\exp ((\theta +2\pi k)J)=\exp (\theta J)$ for every integer $k$, so the Exponential map of a Lie group is surjective onto $SO(2)$ but not injective. Restricting to $-\pi <\theta <\pi$ gives unique angles for every rotation except $R_{\pi }=-I$. The omitted half-turn exposes the cut needed for this single angle chart.

Surjectivity can also fail, even in a connected Lie group. The matrix $A=\mathrm{diag}(-2,-1/2)$ lies in the special linear group $S{L}_2(\mathbb{R})$ but has no real matrix logarithm: any logarithm must commute with $A$, hence preserve each of its distinct real eigenvectors, where exponentiating a real scalar gives a positive number. See the exponential-surjectivity obstruction from distinct negative eigenvalues. These two examples distinguish two different global obstacles: repeated parameters and unreachable elements. The Exponential map of a Lie group remains locally effective in both settings.

The Floquet growth rate measures the exponential trend of a periodic linear differential equation over many coefficient periods. A Floquet multiplier $m$ measures the factor per period $T$; the corresponding rate per unit time iswhere ${\rho }_m$ is a Floquet exponent. Taking the real part removes the ambiguity in the complex logarithm.

For the scalar example under periodic solution, $m=e^{2\pi a}$ and the Floquet growth rate is $a$, independently of $b$. Indeed,The instantaneous logarithmic derivative keeps oscillating even though its average over complete periods is exactly $a$. Sampling at the same phase in each period removes the modulation altogether: $y(nT)=m^{n}y(0)$.

For the general finite-dimensional homogeneous linear differential equation $x^{\prime }=A(t)x$, assume that $A$ is continuous on the real line and $T$-periodic, with $T>0$. Normalize its fundamental matrix by $\Phi (0)=I$ and set $M=\Phi (T)$. The maximal asymptotic exponential rate, using any induced matrix norm, isHere $\rho (M)$ is the spectral radius, $m_j$ are the Floquet multipliers, and invertibility gives $\rho (M)>0$. This concerns growth of the solution operator; it does not say that every initial condition attains the maximum.^[ref]

To prove the formula, write $t=nT+s$ with $0\le s<T$. Periodicity and uniqueness giveBoth $\Phi (s)$ and $\Phi (s{)}^{-1}$ have uniformly bounded matrix norms on a period. Consequently $\Vert \Phi (t)\Vert$ differs from $\Vert M^n\Vert$ by at most fixed multiplicative factors. The spectral radius formula gives $\Vert M^n{\Vert }^{1/n}\to \rho (M)$, proving the claimed limit without choosing a matrix logarithm.

For a particular nonzero initial condition $v$, decompose $v$ into the generalized eigenspaces of $M$ over the complex numbers. The rate of $\Phi (t)v$ is the largest $T^{-1}\log |m_j|$ among components that are nonzero. Generic data have a component in the dominant generalized eigenspaces and attain $\gamma$; data contained in the sum of the subdominant generalized eigenspaces have a smaller rate. For example, with $A=\mathrm{diag}(0.2,-0.3)$, the maximal rate is $0.2$, while $v=(0,1)$ decays at rate $-0.3$. These invariant subspaces describe initial data at phase zero. Their transported versions $E(t)=\Phi (t)E(0)$ can vary with time and satisfy $E(t+T)=E(t)$.

A negative Floquet growth rate allows temporary amplification. For the constant non-normal matrixthe initial condition $v=(0,1)$ has Euclidean norm $e^{-t}\sqrt{100t^2+1}$, which rises from $1$ to about $3.697$ at $t=1$ before eventually decaying. Its exponential rate is still $-1$: the Jordan block contributes a polynomial factor, whose logarithm divided by $t$ tends to zero. At a unit-modulus Floquet multiplier, such polynomial growth can even be unbounded with zero exponential rate; $A=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ gives $e^{tA}v=(t,1)$. Thus zero Floquet growth rate alone proves neither boundedness nor a periodic solution.

For a continuous periodic coefficient matrix $A(t+L)=A(t)$, a fundamental matrix $X$ for $X^{\prime }=A(t)X$, normalized by $X(0)=I$, obeys $X(t+L)=X(t)X(L)$. The invertible matrix $X(L)$ has a complex matrix logarithm; set $B=L^{-1}\log X(L)$ and $P(t)=X(t)e^{-tB}$. Since $X(L)=e^{LB}$ commutes with $B$, direct substitution gives $P(t+L)=P(t)$ and proves the factorization. A real logarithm is not required for this complex representation.

The eigenvalues of $X(L)$ are Floquet multipliers. All solutions are bounded for forward time when every multiplier has modulus at most one and those on the unit circle have no nontrivial Jordan blocks. In a real undamped second-order scalar equation with no first-derivative term, the monodromy determinant is one. Distinct conjugate unit-modulus multipliers give bounded solutions; a nontrivial Jordan block at multiplier $1$ or $-1$ gives an unbounded second solution even when one periodic or antiperiodic solution exists.

The matrix logarithm near the identity maps $Sp(n)=\{A\in U(2n):AJ{A}^t=J\}$ onto an open set of its real compact symplectic Lie algebra $\{X:X^{\ast }=-X,XJ+J{X}^t=0\}$. Its inverse is the matrix exponential. These statements follow by applying the logarithm to $A^{\ast }=A^{-1}$ and $A^t=J^{-1}{A}^{-1}J$, and differentiating $e^{tX}J{e}^{t{X}^t}$. Left translations provide manifold charts at every group element. In block form $X=\left(\begin{array}{cc}P& Q\\ -\bar{Q}& \bar{P}\end{array}\right)$ with $P^{\ast }=-P$ and $Q^t=Q$, giving real dimension $n^2+n(n+1)=n(2n+1)$.

Near the identity, the matrix logarithm of a unitary matrix is skew-Hermitian. Restrict the logarithm's operator norm below $\pi /n$; determinant one then forces its trace to be zero, since $\det U=\exp (\mathrm{tr}\log U)$ and no nonzero multiple of $2\pi i$ fits this bound. The matrix exponential gives the inverse chart on a ball in the displayed real Lie algebra, of dimension $n^2-1$. Left translation supplies compatible charts on the whole special unitary group.

A matrix Lie group is a subgroup of a general linear group that is also a Lie group for its matrix topology. Near the identity, the matrix exponential and matrix logarithm give inverse coordinate charts whenever they are restricted to sufficiently small neighborhoods in the group and its tangent space.

For an odd prime $p$, the matrix logarithm and matrix exponential are inverse maps between $1+p{M}_d({\mathbb{Z}}_p)$ and $p{M}_d({\mathbb{Z}}_p)$. They imply that the $p$th-power map carries $U_m$ bijectively onto $U_{m+1}$. Direct multiplication gives $[U_i,U_j]\subseteq U_{i+j}$, so the group is a powerful pro-p group. The quotient $U_m/U_{m+1}$ is the additive matrix space over ${\mathbb{F}}_p$, with constant index $p^{d^2}$. The $d^2$ matrices $1+p{E}_{ij}$ topologically generate, by successive approximation through these quotients. The lower p-series is exactly $U_m$, so the group is a uniform pro-p group and a torsion-free group. At $p=2$ the first congruence subgroup contains $-I$ and is not uniform.

A uniform pro-p group is a finitely generated powerful pro-p group whose lower p-series $P_1=G$, $P_{i+1}=\overline{P_i^p[P_i,G]}$ has constant successive indices. For odd $p$, the powerful pro-p group condition means $[G,G]\subseteq G^p$, with closed generated powers understood. For $p=2$ it means $[G,G]\subseteq G^4$. For finitely generated powerful pro-p groups, uniformity is equivalent to being a torsion-free group.

It is convenient to use general matrix size $d$, then put $d=p$. Let $U_m=1+p^m{M}_d({\mathbb{Z}}_p)$. The required kernel is $U_1$, the inverse limit of its finite p-group quotients. For odd $p$, the convergent matrix logarithm and matrix exponential seriesare inverse maps between $U_m$ and $p^m{M}_d({\mathbb{Z}}_p)$. The valuation bounds for $A,B\in p{M}_d$ ensure convergence and integrality. They give $\log (u^p)=p\log u$ and hence a bijective power map $U_m\to U_{m+1}$: the unique root of $v\in U_{m+1}$ is $\exp (p^{-1}\log v)$.

Direct matrix multiplication gives $[U_i,U_j]\subseteq U_{i+j}$, so $[U_1,U_1]\subseteq U_2=U_1^p$ and it is a powerful pro-p group. MoreoverThe $d^2$ elements $1+p{E}_{ij}$ topologically generate: their $p^{m-1}$th powers give all matrix basis directions in $U_m/U_{m+1}$, since $(1+p{E}_{ij}{)}^{p^{m-1}}\equiv 1+p^m{E}_{ij}(\mathrm{mod}{p}^{m+1})$. Successive correction of each matrix entry, followed by completeness, approximates every element by words in these generators.

Its lower p-series is exactly $U_m$ and the indices are constant; thus the group is a uniform pro-p group. Alternatively, the logarithm excludes nontrivial p-power torsion. Setting $d=p$ gives a uniform pro-p group kernel of dimension and generator number $p^2$, with indices $p^{p^2}$. This is the odd-prime principal p-adic congruence subgroup is uniform result. The odd-prime condition matters: $-I$ is nontrivial order-two torsion in $1+2M_d({\mathbb{Z}}_2)$.

A Lie group is a group which is a finite-dimensional smooth Hausdorff second-countable manifold, with smooth multiplication and inversion. For the special unitary group, consider the real vector spaceThe diagonal entries of a skew-Hermitian matrix supply $n$ real parameters, and the entries above the diagonal supply $2\left(\genfrac{}{}{0ex}{}{n}{2}\right)$; the trace-zero condition removes one real parameter. Thus ${\dim }_{\mathbb{R}}\mathfrak{su}(n)=n^2-1$.

The matrix facts used to construct charts are these: the matrix exponential is analytic, its derivative at zero is the identity, and it has an analytic inverse matrix logarithm on a sufficiently small neighborhood of the identity. On those neighborhoods $\log (e^X)=X$, $e^{\log U}=U$, and $\det (e^X)=e^{\mathrm{tr}X}$. For a unitary matrix in that neighborhood, the spectral theorem writes $U=W\mathrm{diag}(e^{i{\theta }_j})W^{\ast }$ with principal angles close to zero, and $\log U=W\mathrm{diag}(i{\theta }_j)W^{\ast }$ is skew-Hermitian.

Choose a norm ball of radius $\epsilon <\pi /n$ small enough to lie in the exponential's inverse domain. If $U$ is unitary, $\det U=1$ and $\Vert \log U\Vert <\epsilon$ in operator norm, thenso its trace is zero. Conversely $X\in \mathfrak{su}(n)$ gives $(e^X{)}^{\ast }{e}^X=I$ and $\det (e^X)=1$. Thus the logarithm identifies an open neighborhood $V$ of $I$ in $SU(n)$, with its matrix subspace topology, with the open ball $B_{\epsilon }\subset \mathfrak{su}(n)$. Choosing a real linear identification $\mathfrak{su}(n)\cong {\mathbb{R}}^{n^2-1}$ turns this into a chart.

For every $A\in SU(n)$ take the translated chartTheir overlap maps are $X\mapsto \log (B^{-1}A{e}^X)$ wherever both charts are defined, so they are real analytic. They cover the group; the matrix topology is Hausdorff and second-countable. This constructs the real manifold directly, rather than merely counting equations. Matrix multiplication is polynomial in real/imaginary entries and inversion here is $U\mapsto U^{\ast }$; in the logarithm charts both are smooth. The logarithm charts for the special unitary group therefore proveFor $n=1$ it is the one-point group of dimension zero.

For $n=2$, let the first column be $(a,b)$, of unit norm. Every unit vector orthogonal to it is a scalar of modulus one times $(-\bar{b},\bar{a})$. The determinant of the resulting matrix is that scalar, so determinant one fixes it to $1$. Henceis bijective, with inverse given by taking the first column. The forward map is real polynomial in the four coordinates; composing with the local logarithm charts verifies smoothness into the constructed manifold. The inverse is a smooth matrix-coordinate projection and, in the usual embedded sphere charts, is smooth into $S^3$. Thus $\boxed{SU(2)\cong S^3\text{ by a diffeomorphism}}$, the explicit realization of SU(2) as the three-sphere.

Let $(x^i)$ and $(y^a)$ be overlapping local coordinates at $p$. A tangent vector $v=v_x^i{\partial }_{x^i}$ has components obeying the chain ruleThe Jacobian matrix is evaluated at the same point, and is invertible. For a smooth curve $a$ with $a(0)=p$, define ${\dot{a}}_x^i(0)=\frac{d}{dt}{x}^i(a(t)){|}_{t=0}$. Applying the chain rule to $y(a(t))=y(x(a(t)))$ gives exactly this transformation law. Thus these components determine a well-defined velocity vector $\dot{a}(0)\in T_{p}M$, independent of the chart. Equivalently it is the derivation $f\mapsto \frac{d}{dt}f(a(t)){|}_0$ on smooth functions. The two curve symbols used in the printed first paragraph refer to this same curve.

For real scalars $\lambda ,\mu$, considerFor sufficiently small positive and negative $t$, both factors belong to $G$, so $c$ is a curve in $G$ with $c(0)=I$. The standard matrix exponential properties used here are $\exp (0)=I$, $\frac{d}{dt}\exp (tB){|}_0=B$, and $\exp (-tB)=\exp (tB{)}^{-1}$. The product rule gives $c^{\prime }(0)=\lambda B_1+\mu B_2$. Every such linear combination is therefore a velocity at the identity.

More generally, the inclusion $j:G\hookrightarrow GL(n,\mathbb{R})$ is an embedding, so its tangent map at $I$ is injective and linear. The general linear group is open in the vector space of all matrices; its tangent space at $I$ is consequently $M_n(\mathbb{R})$. We identifyEvery tangent vector is the velocity of a curve in a manifold chart, so this subspace is precisely the set of the corresponding matrix velocities. The product construction is consistent with its vector-space operations.

Now use the stated logarithm-chart assumption. Near zero and $I$, the matrix exponential and matrix logarithm are inverse smooth maps, with $d{\exp }_0=d{\log }_I=\mathrm{id}$. Hence for any $B\in \mathfrak{g}$, $\exp (tB)$ belongs to $G$ for small $t$: $tB$ lies in the neighborhood which is the image of the logarithm. For $B_1,B_2\in \mathfrak{g}$, form the group commutatorMultiplying the exponential expansions $e^{tB}=I+tB+\frac{1}{2}{t}^2{B}^2+O(t^3)$ givesThe convergent local series $\log (I+Z)=Z-\frac{1}{2}{Z}^2+\cdots$ therefore gives $\log C(t)=t^2[B_1,B_2]+O(t^3)$. For small nonzero $t$, the matrix $t^{-2}\log C(t)$ belongs to the linear subspace $\mathfrak{g}$. A finite-dimensional linear subspace is closed, so taking its limit provesThis is the matrix commutator from a logarithm chart argument. It identifies $\mathfrak{g}$ as the Lie algebra of a matrix Lie group, with the matrix commutator as its Lie bracket. In particular, no potentially nonsmooth square-root change of parameter is needed to turn the second-order commutator into a velocity.

The proof of the linearity of a uniform pro-p group has three distinct steps: pass to a finite-dimensional Lie algebra, integrate a faithful representation on a sufficiently small open subgroup, and extend it to the entire group. Working first on a small subgroup is essential because a faithful Lie algebra representation need not send the original integral lattice into the convergence domain of the matrix exponential.

Let $G$ be a uniform pro-p group and let $L_G$ be its intrinsic powerful Lie lattice. Thenis a finite-dimensional Lie algebra over a field of characteristic zero. The Ado theorem, the classical Lie-theoretic ingredient in this proof, supplies a faithful finite-dimensional Lie algebra representationOne must use a faithful representation rather than just the Adjoint representation, which annihilates the center of a Lie algebra.

Choose a p-adic lattice $\Lambda \subset V$. Since $L_G$ is finitely generated as a module, the entries of the representing matrices of a module basis have bounded denominators. Hence for sufficiently large $k$,For odd $p$, the matrix exponential and p-adic logarithm converge and are inverse between $p\mathrm{End}(\Lambda )$ and $1+p\mathrm{End}(\Lambda )$. Put $H=G^{p^k}$; under the uniform pro-p group and powerful Lie lattice correspondence it is the BCH group on $p^k{L}_G$. DefineHere ${\exp }_G$ denotes the inverse of the intrinsic logarithm identification, not an assumed matrix representation of $G$. The Lie algebra homomorphism $\rho$ respects the convergent Baker--Campbell--Hausdorff formula, soThus $\tau$ is a continuous group homomorphism. If $\tau ({\exp }_{G}X)=1$, taking the matrix logarithm gives $\rho (X)=0$, whence $X=0$. Therefore $\tau$ is faithful and all its matrices preserve $\Lambda$. This proves linearity of the open normal subgroup $H$.

To finish, use the extension of a faithful representation from an open normal subgroup. Choose representatives $t_1=1,t_2,\dots ,t_r$ of the finite left cosets $G/H$, where $r=[G:H]$. On the finite free module $W={⨁}_{j=1}^r\Lambda$, define the action of $g\in G$ by writing uniquelyand mapping the $j$-th block to the $i$-th block by $\tau (h_{ij})$. If $g^{\prime }{t}_j=t_{\ell }{h}^{\prime }$ and $g{t}_{\ell }=t_{i}h$, then $g{g}^{\prime }{t}_j=t_{i}h{h}^{\prime }$; this proves the representation identity because $\tau (h{h}^{\prime })=\tau (h)\tau (h^{\prime })$. Each block matrix is invertible over ${\mathbb{Z}}_p$, and the map is continuous because the finite coset permutation is locally constant and the remaining matrix entries are obtained continuously from $\tau$.

If $g\notin H$, its permutation of $G/H$ moves the identity coset, so its block representation is not the identity. If $g\in H$ and its block representation is the identity, the block corresponding to $t_1=1$ is $\tau (g)=1$, and faithfulness of $\tau$ gives $g=1$. Thus this induced representation is faithful. Choosing a basis of $W$ provesThe continuous injection is a topological embedding, since $G$ is compact and the target is Hausdorff. The trivial group is immediate. For uniform groups at $p=2$, the same proof uses the usual stronger powerfulness condition and the convergence ball $4\mathrm{End}(\Lambda )$ instead; the preceding questions concern odd $p$.

For probability distributions $p$ and $q$ on a finite alphabet, their classical relative entropy in bits isA zero-$p$ term contributes zero, including when $q(x)=0$; a term with $p(x)>0$ and $q(x)=0$ makes the result $+\infty$. In that infinite case nonnegativity is immediate. Otherwise $q(x)>0$ on the support $J_p$ of $p$. For every $u>0$, $\ln u\le u-1$: the function $u-1-\ln u$ has derivative $1-1/u$ and its minimum zero at $u=1$. Apply this with $u=q(x)/p(x)$ to obtainThe last inequality uses normalization of $q$. Thus $\boxed{D(p\Vert q)\ge 0}$. Equality requires $q(x)=p(x)$ wherever $p(x)>0$ and no $q$-mass outside that support, hence occurs precisely when $p=q$.

For density matrices, usewhen the support of a positive operator $\rho$ is contained in that of $\sigma$, and set it to $+\infty$ otherwise. The data-processing inequality for quantum relative entropy states that for any quantum channel $\Phi$,Thus the relative entropy of the two output states cannot exceed that of the inputs.

For the states with a classical flag, omit indices of weight zero. On a positive-weight block, the matrix logarithms obeyon the relevant supports. The logarithms of $p_i$ cancel. Taking the trace block by block therefore gives the relative entropy of classically flagged states:If support inclusion fails in a positive-weight block, both sides are infinite. A zero-weight block makes no contribution, even if its unweighted relative entropy is infinite.

Tracing out the flag is a quantum channel with operators $A_i=⟨i|\otimes I$, since ${\sum }_i{A}_i^{†}{A}_i=I$. It sends the two flagged states to ${\sum }_i{p}_i{\rho }_i$ and ${\sum }_i{p}_i{\sigma }_i$. Applying the data-processing inequality for quantum relative entropy to this partial trace proves joint convexity of quantum relative entropy:

For a bipartite state, set ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$ and ${\rho }_B={\mathrm{Tr}}_A{\rho }_{AB}$. Its quantum mutual information iswhere $S(\tau )=-\mathrm{Tr}\tau {\log }_2\tau$ is the Von Neumann entropy. The support of ${\rho }_{AB}$ lies in $\mathrm{supp}{\rho }_A\otimes \mathrm{supp}{\rho }_B$. To justify this when marginals are singular, let $Q_A$ project onto $\ker {\rho }_A$. Then $\mathrm{Tr}[(Q_A\otimes I){\rho }_{AB}]=\mathrm{Tr}(Q_A{\rho }_A)=0$; positivity implies $(Q_A\otimes I){\rho }_{AB}=0$. Applying the same argument on $B$ proves the claimed support inclusion.

On that support, product eigenvectors showThe definition of the partial trace therefore givesConsequently $\boxed{S(A:B)=S({\rho }_{AB}\Vert {\rho }_A\otimes {\rho }_B)}$.

For completeness, nonnegativity of quantum relative entropy also follows from the classical proof above. Diagonalize $\rho$ in a basis $|j⟩$, with eigenvalues $r_j$, and put $q_j=⟨j|\sigma |j⟩$. These are two classical probability distributions. Scalar concavity of the logarithm applied to the spectral weights of $\sigma$ gives $⟨j|{\log }_2\sigma |j⟩\le {\log }_2{q}_j$ whenever $r_j>0$ and support inclusion holds. HenceThe support-failure case is again infinite. Applying this to the product of the marginals proves $\boxed{S(A:B)\ge 0}$ without requiring an additional entropy inequality.

Fix the orientation of the Berezin integral by defining its measure as the linear functional that extracts the coefficient of ${\prod }_{i=1}^N({\theta }_i{\bar{\theta }}_i)$; all monomials of lower degree integrate to zero. In particular $\int \mathcal{D}(\bar{\theta },\theta ){\prod }_i({\theta }_i{\bar{\theta }}_i)=1$. This specifies the sign convention in the notation $d^N\bar{\theta }{d}^N\theta$.

The Grassmann change-of-variables formula givesFor a unitary matrix, $\det U\det U^{\ast }=1$, so the combined measure is invariant. Diagonalize the Hermitian matrix $B$ by a unitary change of variables. If its eigenvalues are $b_i$, the integrand factors as ${\prod }_i(1-b_i{\bar{\theta }}_i{\theta }_i)={\prod }_i(1+b_i{\theta }_i{\bar{\theta }}_i)$. Extracting its top coefficient proves the Grassmann Gaussian integralThe determinant identity also holds without Hermiticity by direct expansion; Hermiticity makes the diagonalization proof immediate.

Let $C(x)={\sum }_k{x}_k{C}^{(k)}$. Integrating out the Grassmann variables first gives the exact polynomial expressionThe requested factorization requires $B$ to be invertible, an assumption omitted from the question. Under it, use the formal perturbative matrix logarithm near $g=0$:ConsequentlyExpanding $C(x)$ gives exactly the displayed coefficients $V_{k_1\dots k_m}^{(m)}=(m-1)!\mathrm{Tr}[B^{-1}{C}^{(k_1)}\cdots B^{-1}{C}^{(k_m)}]$, and hence $I(g)=\det B\int d^{N}x{e}^{-x^{T}Ax/2-V_{\mathrm{eff}}(x)}$. The logarithmic representation is a formal series, not a globally convergent logarithm for every $x$: the determinant can have zeros. The original determinant integral remains well-defined for singular $B$, but $I(0)=0$ and the requested normalized expansion then cannot be formed.

There is a second convention to specify. The $x_i$ are real, so a complex Hermitian $A$ gives $x^{T}Ax=x^T{A}_{R}x$, where $A_R=\mathrm{Re}A=(A+A^T)/2$ is real symmetric positive definite. Its Gaussian covariance matrix is $A_R^{-1}$. It equals $A^{-1}$ only if $A$ is real symmetric. ThusThe effective Feynman rules are: a bosonic contraction joins indices $k,\ell$ with $(A_R^{-1}{)}_{k\ell }$; an $m$-leg vertex carries $-g^m\mathrm{Sym}{V}_{k_1\dots k_m}^{(m)}$; sum all closed vacuum diagrams with their symmetry factors, including disconnected diagrams. The symmetrization is necessary because the $x_k$ commute: the trace tensor is cyclically invariant but need not be fully symmetric. The minus sign comes from expanding $e^{-V_{\mathrm{eff}}}$. Taking the logarithm of the normalized integral instead keeps only connected vacuum diagrams.

Equivalently, retain the fermions: an oriented contraction is $(B^{-1}{)}_{ij}=⟨{\theta }_i{\bar{\theta }}_j⟩$, the mixed vertex has weight $+g{C}_{ij}^{(k)}$, and every closed fermion loop gives a minus sign. The same bosonic covariance joins its $x$ legs. Expanding these contractions recovers the trace vertices above. This is bosonic trace vertices from a finite fermionic determinant.

A Lie group is a finite-dimensional smooth manifold with a group structure for which multiplication and inversion are smooth. The group defined here is the compact symplectic group $Sp(n)=U(2n)\cap Sp(2n,\mathbb{C})$, rather than the full complex symplectic group. Also, the displayed matrix expression requires $B$ to be $2n\times 2n$; the printed $n\times n$ size is incompatible with $J$.

Since $J^2=-I$, we have $J^{-1}=-J$. The matrix exponential commutes with conjugation, as follows term by term from its absolutely convergent power series. ConsequentlyTo construct logarithm charts for the compact symplectic group, consider the real vector spaceDifferentiating the defining identities at $I$ gives precisely these conditions. Conversely if $X\in \mathfrak{k}$, $e^X$ is unitary andso $e^X\in Sp(n)$. These are the infinitesimal conditions of the compact symplectic Lie algebra.

Near $I$, the convergent matrix logarithm series $\log (I+C)={\sum }_{m\ge 1}(-1{)}^{m+1}{C}^m/m$ is smooth and inverse to the matrix exponential near zero. These local inverses respect transpose, conjugate transpose, and conjugation; also $\log (A^{-1})=-\log A$ when both matrices are sufficiently close to $I$. Shrink their neighborhoods accordingly. If $A\in Sp(n)$ there, unitarity givesThe symplectic identity is equivalent to $A^t=J^{-1}{A}^{-1}J$, so, putting $X=\log A$, it gives $X^t=-J^{-1}XJ=JXJ$, equivalently $XJ+J{X}^t=0$. Thus this local logarithm restricts to a bijection between a neighborhood of $I$ in $Sp(n)$ and an open neighborhood of zero in $\mathfrak{k}$. Its inverse is the restricted matrix exponential. These restrictions are manifold charts; left multiplication translates them to every $A_0\in Sp(n)$ via $A\mapsto \log (A_0^{-1}A)$. The chart overlaps are smooth compositions of multiplication, exponential and logarithm. The subspace topology is Hausdorff and second countable, inherited from the finite-dimensional matrix space, so these charts give a smooth manifold.

Closure under products and inverses follows from $AJ{A}^t=J$ and unitarity. Matrix multiplication is polynomial in real and imaginary entries, and inversion on the unitary group is $A\mapsto A^{\ast }$, a real linear operation. Their restrictions are smooth in the charts just constructed. Hence $Sp(n)$ is a Lie group without needing a closed-subgroup theorem.

Write $X$ in $n\times n$ blocks. The two infinitesimal conditions giveThe skew-Hermitian matrix $P$ has $n^2$ real parameters; the complex symmetric matrix $Q$ has $n(n+1)/2$ complex parameters, hence $n(n+1)$ real parameters. ThereforeFor $n=1$, any two-by-two matrix satisfies $AJ{A}^t=(\det A)J$, so the group is $SU(2)$. Explicitly,The rows are orthonormal and the determinant is one; conversely unitarity and determinant one force this form. This is the SU(2) as the three-sphere parametrization. The map and its inverse, extraction of the first row, are smooth. Thus $Sp(1)$ is diffeomorphic to $S^3$.

A real Lie group is a group $G$ equipped with a finite-dimensional real smooth manifold structure, conventionally Hausdorff and second countable, for which multiplication $G\times G\to G$ and inversion $G\to G$ are smooth.

Its tangent space at the identity $e$ is defined by smooth curves through $e$: two curves represent the same tangent vector when their derivatives in a local chart agree at zero. For $X\in T_{e}G$, left translation $L_g(h)=gh$ determines the left-invariant vector fieldThe commutator of these derivations on smooth functions is another left-invariant vector field, so define the Lie bracketThis construction supplies the Lie algebra structure. In a Matrix Lie group, $X^L(g)=gX$ and differentiating the two fields gives $[X,Y]=XY-YX$.

For the special linear group, differentiate the determinant along a curve $g(t)$ through $I$. Expansion of the determinant, or its differential $D{\det }_A(H)=\det (A)\mathrm{tr}(A^{-1}H)$, givesThe determinant differential is surjective at $I$, so the level set $\det =1$ has tangent space equal to its differential's kernel. Equivalently, every tangent matrix has zero trace, and every zero-trace matrix supplies a curve $e^{tX}$ of determinant $e^{t\mathrm{tr}X}=1$. Hencewith the matrix commutator bracket. The same calculation over $\mathbb{C}$ gives the complex special linear Lie algebra ${\mathfrak{sl}}_n(\mathbb{C})$; as a real group, the complex group has that space viewed as a real Lie algebra.

The matrix exponential is the everywhere-convergent serieswhose inverse matrix is $\exp (-X)$. The matrix logarithm is locally defined near $I$ byThese maps are inverse on suitable neighbourhoods of zero and $I$, giving a logarithmic chart of a matrix Lie group. A logarithm is not a globally single-valued inverse of the exponential.

Every invertible complex matrix nevertheless has at least one matrix logarithm. Put it in Jordan normal form. For a block $J=\lambda I+N$, $\lambda \ne 0$ and $N^s=0$, choose any complex scalar logarithm $\ell$ of $\lambda$ and setThe finite logarithm and exponential identities in the nilpotent variable give $e^L=\lambda (I+N/\lambda )=J$. Combine the blocks and conjugate back. Thus the exponential map is surjective on ${\mathrm{GL}}_n(\mathbb{C})$, by existence of a logarithm for every invertible complex matrix.

A connected counterexample is ${\mathrm{SL}}_2(\mathbb{R})$. It is connected: polar decomposition of an invertible real matrix writes each element as $KP$, with $K\in \mathrm{SO}(2)$ and $P$ positive definite symmetric of determinant one; $\mathrm{SO}(2)$ is connected and $P$ is connected to $I$ through $P^t$.

But $g=\mathrm{diag}(-2,-1/2)$ belongs to this group and is not $e^X$ for any real $X$. Such an $X$ would commute with $e^X=g$. Since $g$ has two distinct real eigenvalues, direct commutation makes $X$ a real diagonal matrix. Its exponential has positive diagonal entries, a contradiction. ThereforeThis is an exponential-surjectivity obstruction from distinct negative eigenvalues; connectedness does not eliminate the obstruction.

A real Lie group is a finite-dimensional real smooth manifold with a group structure whose multiplication and inversion are smooth. Its tangent space at the identity element $e$ is the vector space $\mathfrak{g}=T_{e}G$ of velocities of smooth curves through $e$. For the general linear group ${\mathrm{GL}}_n(\mathbb{R})$ it identifies with $M_n(\mathbb{R})$. The matrix exponential and local matrix logarithm areThe exponential is globally defined, and each matrix $\exp X$ is invertible with inverse $\exp (-X)$. The logarithm here is a local map near $I$; it is not a globally defined inverse on every real invertible matrix. The differential of $\exp$ at zero is the identity, and the two power series are inverse locally.