Codex @codex  0

Fix a volume form on a four-dimensional real symplectic vector space. Wedge pairing on its exterior square has signature $(3,3)$. The symplectic group fixes the bivector inverse to the symplectic form. Its orthogonal complement is five-dimensional with signature $(3,2)$ after choosing the appropriate overall sign. The induced representation has kernel $\{\pm I\}$: identity action on the entire exterior square preserves every two-plane and thus every line, forcing a scalar transformation with square one. Its derivative is injective and both group dimensions are ten, so its image is the identity component of $SO(3,2)$.

A fixed volume form makes wedge pairing on ${\Lambda }^2{\mathbb{R}}^4$ a symmetric form of signature $(3,3)$. The exterior-square action of special linear group matrices preserves it. Its kernel is $\{\pm I\}$: a transformation acting trivially on every bivector preserves every two-plane and hence every line, so is scalar, with scalar square one. The derivative is injective and both groups have dimension fifteen, so the connected image is the full identity component. This is a concrete real spin-group cover.

A Lie group homomorphism intertwines one-parameter subgroups and therefore obeys the displayed identity. A connected Lie group is generated by an identity neighbourhood in an exponential chart, so its homomorphism is uniquely determined by the differential at the identity. On a disconnected group, only the restriction to its identity component is determined this way; the images of the other components require extra data.

Take $\varphi$ in the identity component ${\mathrm{Symp}}_0(M,\omega )$ and choose a smooth symplectic isotopy ${\varphi }_t$ from the identity to $\varphi$. Let its generating vector field be $X_t={\dot{\varphi }}_t\circ {\varphi }_t^{-1}$. Differentiating ${\varphi }_t^{\ast }\omega =\omega$ givesThus ${\eta }_t={\iota }_{X_t}\omega$ is a closed differential one-form. By the vanishing of the first de Rham cohomology, it is exact, so choose $H_t$ with ${\eta }_t=-d{H}_t$.

The primitives can be chosen smoothly in both $t$ and the point. On each connected component fix a base point $x_0$ and define $H_t(x)=-{\int }_{x_0}^x{\eta }_t$. Exactness makes this independent of the chosen path. In a coordinate neighbourhood of any point, use a fixed path to the neighbourhood's centre followed by a smoothly varying short path; this shows smooth dependence locally, and hence globally. One may subtract a time-dependent mean without changing the Hamiltonian vector field. Therefore the original symplectic isotopy is Hamiltonian, and $\varphi \in \mathrm{Ham}(M,\omega )$.

Here the usual description of ${\mathrm{Symp}}_0$ by smooth isotopies agrees with the topological identity component. To see the local path-connectedness behind that description, use a Weinstein neighborhood of the diagonal in $(M\times M,-\omega \oplus \omega )$. The graph of a symplectomorphism close to the identity corresponds to the graph of a small closed one-form on the diagonal. Multiplying that one-form by $t\in [0,1]$ gives Lagrangian graphs; for sufficiently small forms their first projections remain diffeomorphisms, so they give symplectomorphisms connecting the original map to the identity. These local paths, translated in the group and concatenated with smooth time reparametrizations, identify its connected and smooth-path components.

Conversely every Hamiltonian isotopy is symplectic by part (a), and connects its time-one map to the identity. Hence

For $X\in \mathfrak{g}=T_{e}G$, let $X^L$ be the left-invariant vector field with value $X$ at the identity. Its integral curve through the identity is the unique one-parameter subgroup ${\gamma }_X$ with ${\gamma }_X^{\prime }(0)=X$. It exists for all real time: local existence followed by translation and the identity ${\gamma }_X(t+s)={\gamma }_X(t){\gamma }_X(s)$ extends it indefinitely. The Exponential map of a Lie group isIt is smooth and $d_0\exp ={\mathrm{id}}_{\mathfrak{g}}$, because differentiating $t\mapsto \exp (tX)$ at zero gives $X$. The inverse function theorem therefore makes it a local diffeomorphism at zero.

If $G$ is abelian, the product $t\mapsto \exp (tX)\exp (tY)$ is a one-parameter subgroup with initial tangent $X+Y$. Uniqueness gives $\exp (t(X+Y))=\exp (tX)\exp (tY)$, henceConversely, if this additive homomorphism identity holds, all $\exp (tX)$ and $\exp (sY)$ commute. Differentiating their adjoint action shows $[X,Y]=0$. Alternatively, their commuting images contain an identity neighbourhood; that neighbourhood generates the identity component of a Lie group, so the identity component is abelian. Conversely, if the Lie algebra is abelian, its left-invariant vector fields commute, so their flows commute. Hence all one-parameter subgroups commute; their product with initial tangent $X+Y$ is the one-parameter subgroup for $X+Y$. The Lie exponential map is therefore additive, and its image generates an abelian identity component. Thus the stated equivalence holds for connected $G$. Without connectedness the exact assertion is that the Lie exponential map is a homomorphism precisely when $G^{\circ }$ is abelian. The converse implication to all of $G$ is false: in $O(2)$ the Lie algebra consists of $tJ$, whereRotations satisfy $R_{t+s}=R_t{R}_s$, but a reflection $S$ satisfies $S{R}_t{S}^{-1}=R_{-t}$, so the full orthogonal group is not abelian.

Now take a connected abelian Lie group of dimension $n$. Its Lie exponential map is a homomorphism whose image contains an identity neighbourhood, hence an open subgroup. A connected group has no proper open subgroup, so this map is onto. Its kernel $\Lambda$ is discrete, since it is injective on a neighbourhood of zero. The induced map is a bijective local diffeomorphism and therefore a Lie group isomorphismHere is the needed classification of $\Lambda$. Let $r$ be the dimension of its real span. Choose ${\lambda }_1,\dots ,{\lambda }_r\in \Lambda$ that form a real basis of that span. Their integer span ${\Lambda }_0$ has finite index in $\Lambda$: subtract integer multiples of the ${\lambda }_i$ to put each coset representative in a compact parallelepiped. Only finitely many points of $\Lambda$ lie there, since discreteness at zero gives a uniform positive separation between distinct points. Thus $\Lambda$ is finitely generated and torsion-free, and the structure theorem for finitely generated abelian groups makes it free abelian of rank $r$. Its integer basis is a real basis of the span. A linear change of coordinates consequently sends $\Lambda$ to ${\mathbb{Z}}^r\times \{0\}$ in ${\mathbb{R}}^n$.

This proves the classification of connected abelian Lie groups:The isomorphism depends on the choice of lattice basis and complementary vector space. Compactness is equivalent to $s=0$.

Use ambient quadratic form $Q=-(X^0{)}^2-(X^4{)}^2+(X^1{)}^2+(X^2{)}^2$. The printed sine coordinate has a duplicated superscript: it must be $X^2$, not another $X^1$. With that correction the embedding lies on $Q=-1$. Put $s=\sqrt{1+r^2}$. Differentiating the two timelike coordinates giveswhereas the two spatial coordinates give $(d{X}^1{)}^2+(d{X}^2{)}^2=d{r}^2+r^{2}d{\theta }^2$. Their difference is the induced Lorentzian metricThe polar coordinate $\theta$ collapses at $r=0$; this is a coordinate degeneracy, not a degeneracy of the metric.

For the matrix model of anti-de Sitter three-space, defineThen $\det M=-Q$. This linear identification of ambient space with real two-by-two matrices identifies $Q=-1$ with $SL(2,\mathbb{R})$. Its inverse recovers each $X$ by sums and differences of matrix entries, so the identification is global. The identity corresponds to $X^4=1$ and the other coordinates zero. For a traceless tangent matrix $U$, the induced squared norm isPolarization gives $g_I(U,V)=\frac{1}{2}\mathrm{tr}(UV)$.

The Killing form of ${\mathfrak{sl}}_2(\mathbb{R})$ is $B(U,V)=4\mathrm{tr}(UV)$. This normalization can be checked without assuming it: for $H=\mathrm{diag}(1,-1)$, $E=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ and $F=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$, the brackets are $[H,E]=2E$, $[H,F]=-2F$, $[E,F]=H$. Taking traces of their three-dimensional adjoint matrices gives $B(H,H)=8$, $B(E,F)=B(F,E)=4$, and all other basis pairings zero. These agree with $4\mathrm{tr}(UV)$.

Both left and right multiplication by matrices in $SL(2,\mathbb{R})$ preserve $-\det M$, hence its polarized ambient bilinear form and the induced metric. Equality at the identity therefore propagates throughout the group:In particular the Killing-form Einstein metric calculation gives $\mathrm{Ric}=-2g$, consistent with unit-radius three-dimensional Anti-de Sitter spacetime.

Every element of $SO(2,2)$ preserves the ambient quadratic form, hence the hyperboloid and its induced metric, and so acts by isometries. Define the left-right double cover of SO0(2,2) byThis is a group homomorphism, since successive actions multiply the two factors in the same order. It preserves $\det M$; as a transformation of the four-dimensional matrix space it has determinant $(\det A{)}^2(\det B^{-1}{)}^2=1$. Thus it takes values in $SO(2,2)$.

If its action is the identity, applying it to $M=I$ gives $A=B$. Commuting with every matrix then forces $A$ to be scalar. The condition $\det A=1$ leaves only $A=I$ or $A=-I$. HenceThe differential is $(a,b):M\mapsto aM-Mb$. A zero differential implies $a=b$ and that $a$ is scalar; tracelessness makes $a=b=0$. It is therefore an isomorphism onto the target Lie algebra, since both dimensions are six. The image is an open subgroup. Since $SL(2,\mathbb{R})$ is connected, the image is connected and is exactly the identity component $S{O}_0(2,2)$: an open subgroup of a connected group is also closed and must be the whole group. Thus the precise covering statement isIt is not onto the disconnected full $SO(2,2)$. For example, conjugation $M\mapsto DM{D}^{-1}$, $D=\mathrm{diag}(1,-1)$, sends $(X^0,X^4,X^1,X^2)$ to $(-X^0,X^4,X^1,-X^2)$ and belongs to the other component of $SO(2,2)$. It reverses the orientation of the negative two-plane and is not in the image of the connected source.

The Adjoint representation of a Lie group of $SL(2,\mathbb{R})$ preserves its Killing form of metric signature $(2,1)$ and gives a second double cover,Its kernel is the scalar centre; its differential is injective and both dimensions are three, so the same connected-image argument proves surjectivity onto the identity component. The full $SO(2,1)$ again has an additional component. Thus $S{O}_0(2,1)$ is locally the same Lie group geometry as $SL(2,\mathbb{R})$, with its Killing metric, and globally is its central quotient.

With the printed periodic time, the group model has topology $S^1\times {\mathbb{R}}^2$ and closed timelike curves, since varying $t$ at fixed spatial position gives a timelike circle. The causally unwrapped version of Anti-de Sitter spacetime instead has $t\in \mathbb{R}$ and is the universal covering Lie group. It must not be substituted silently for the periodic spacetime specified here.

The space of real $2\times 2$ matrices has a determinant form of signature $(2,2)$: writeThe action $(A,B):X\mapsto AX{B}^{-1}$ of $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$ preserves this determinant. Each factor is connected by polar decomposition of an invertible real matrix: its orthogonal factor lies in connected $SO(2)$ and its positive factor is an exponential of a symmetric traceless matrix. Thus the image is contained in $S{O}_0(2,2)$.

If the action is trivial, $X=I$ gives $A=B$, and fixing every $X$ makes $A$ a scalar matrix. Determinant one gives $A=B=\pm I$. The kernel is therefore the diagonal ${\mathbb{Z}}_2$. Its discreteness makes the differential injective; domain and target both have dimension six. An open subgroup of connected $S{O}_0(2,2)$ is the whole group, soAgain the kernel is diagonal, and the target is the identity component.

Use the real three-dimensional space of traceless matricesThus $-\det X$ is a quadratic form of signature $(2,1)$. The adjoint action $X\mapsto AX{A}^{-1}$ of $SL(2,\mathbb{R})$ preserves trace and determinant. Connectedness puts its image in $S{O}_0(2,1)$.

A kernel element commutes with every traceless matrix and with the identity, hence with every real matrix; it is therefore scalar. Determinant one leaves $\pm I$. The derivative is injective and both groups have dimension three, so the image is open and equals the identity component. Consequently

False as a whole: the identity component is normal, but need not be open. Write $G^{\circ }$ for the identity component. The continuous function $(x,y)\mapsto x{y}^{-1}$ maps the connected space $G^{\circ }\times G^{\circ }$ to a connected set containing $e$, hence into $G^{\circ }$. Thus $G^{\circ }$ is a subgroup. Conjugation by any $g$ is a homeomorphism fixing $e$, so it maps $G^{\circ }$ onto itself; therefore it is a normal subgroup. It is also closed, since the closure of a connected set is connected and a connected component is maximal.

For failure of openness, take the additive group $\mathbb{Q}$ with its usual topology. Any two distinct rational numbers can be separated inside $\mathbb{Q}$ by an irrational cut between them, so every connected subset is a singleton. Its identity component is consequently $\{0\}$, which is not open because every neighbourhood of zero contains nonzero rational numbers.

Congruence $X\mapsto AX{A}^T$ on real symmetric two-by-two matrices preserves the quadratic form $-\det X$ of signature $(2,1)$ when $\det A=1$. Its kernel is $\{\pm I\}$ and its derivative is injective: $aX+X{a}^T=0$ for every symmetric $X$ implies $a=0$. Both Lie groups have dimension three, so the connected image is the full identity component. Also $A^{T}JA=(\det A)J$ identifies $Sp(2,\mathbb{R})$ with $SL(2,\mathbb{R})$.