用户: Solution/ 习题: 规范场/Example Sheet 1

范围包括微分流形、李群、李代数、表示.

Question 1. Let $M$ and $N$ be smooth manifolds and $\varphi :M\to N$ a smooth map.

•	Suppose $X$ and $Y$ are vector fields on $M$ and $N$ respectively. If $Y$ is $\varphi$-related to $X$ (i.e. $Y_{\varphi (p)}=(D_p\varphi )(X_p)$,$\forall p\in M$), prove that $$(L_{Y}f)\circ \varphi =L_X(f\circ \varphi ),\forall f\in C^{\infty }(N).$$
•	If $X^{\prime }$ is $\varphi$-related to $Y$, prove that $[X^{\prime },Y^{\prime }]$ is $\varphi$-related to $[X,Y]$.

Solution:

If $Y_{\varphi (p)}=(D_p\varphi )X_p,\forall p\in M$, then$$\begin{array}{rl}(L_{Y}f)\circ \varphi (p)& =(D_{\varphi (p)}f)(Y_{\varphi (p)})=(D_{\varphi (p)}f)(D_p\varphi )(X_p)\\ & =((D_p\varphi )X_p)(f)=L_X(f\circ \varphi )(p).\end{array}$$

Hence $X^{\prime }{Y}^{\prime }(f\circ \varphi )=X^{\prime }((Yf)\circ \varphi )=(XYf)\circ \varphi$; similarly, $Y^{\prime }{X}^{\prime }(f\circ \varphi )=(YXf)\circ \varphi$. Therefore,$$\begin{array}{rl}[X^{\prime },Y^{\prime }](f\circ \varphi )& =X^{\prime }{Y}^{\prime }(f\circ \varphi )-Y^{\prime }{X}^{\prime }(f\circ \varphi )\\ & =(XYf)\circ \varphi -(YXf)\circ \varphi \\ & =([X,Y]f)\circ \varphi .\end{array}$$

Question 2. Show that

•	$\mathrm{GL}(n,\mathbb{R})$ is a non-compact Lie group of $n^2$ dimension;
•	$\mathrm{GL}(n,\mathbb{C})$ is a non-compact Lie group of $2n^2$ dimension;
•	both of them are non-abelian for $n\ge 2$.

Solution:

Let $M(n,\mathbb{R})$ denote the set of $n\times n$ matrices with real entries. Because it is a real vector space of dimension $n^2$ under matrix addition and scalar multiplication, $M(n,\mathbb{R})$ is a smooth $n^2$-dimensional smooth manifold. Now $\mathrm{GL}(n,\mathbb{R})$ is an open subset of $M(n,\mathbb{R})$, namely the set where $\det :M(n,\mathbb{R})\to \mathbb{R}$ is nonzero, hence $\mathrm{GL}(n,\mathbb{R})$ is a smooth $n^2$-dimensional submanifold of $M(n,\mathbb{R})$. Moreover, for all $x\in \mathrm{GL}(n,\mathbb{R})$, $\lambda x\in \mathrm{GL}(n,\mathbb{R})(\lambda \in \mathbb{R}-\{0\})$. Therefore $\mathrm{GL}(n,\mathbb{R})$ is not bounded, hence non-compact.

Next we prove that $\mathrm{GL}(n,\mathbb{R})$ admits a Lie group structure. Multiplication is smooth because the matrix entries of a product matrix $AB$ are polynomials in the entries of $A$ and $B$. Inversion is smooth by Cramer’s Rule. When $n\ge 2$, $AB\ne BA$, hence $\mathrm{GL}(n,\mathbb{R})$ is not Abelian when $n\ge 2$.

The case for $\mathrm{GL}(n,\mathbb{C})$ is similar.

Question 3. Find an explicit Lie group homomorphism $$\varphi :SU(3)\times SU(2)\times U(1)\to SU(5)$$with discrete kernel.

Solution:

Consider the homomorphism$$\begin{array}{r}\varphi :SU(3)\times SU(2)\times U(1)\to SU(5)\\ (U_3,U_2,e^{i\theta })\mapsto \left(\begin{array}{cc}{e}^{-2i\theta }{U}_3& 0\\ 0& e^{3i\theta }{U}_2\end{array}\right).\end{array}$$

•

$\varphi$ is indeed a Lie group homomorphism. $\varphi$ is clearly a group homomorphism; to prove $\varphi$ is smooth, note just $$\varphi (U_3,U_2,e^{i\theta })=\mathrm{diag}\{e^{-2i\theta },e^{-2i\theta },e^{-2i\theta },e^{3i\theta },e^{3i\theta }\}\left(\begin{array}{cc}{U}_3& 0\\ 0& U_2\end{array}\right).$$ The two terms of the left hand side of the above equation are embeddings of $U(1)$ and $SU(3)\times SU(2)$ into $SU(5)$, respectively. Hence they are smooth; but also product operation in Lie group is smooth, so $\varphi$ is smooth.

•

The kernel of $\varphi$ is discrete. If $(U_3,U_2,e^{i\theta })\in \mathrm{Ker}\varphi$, then $U_3=\mathrm{diag}\{e^{2i\theta },e^{2i\theta },e^{2i\theta }\}$, $U_2=\mathrm{diag}\{e^{-3i\theta },e^{-3i\theta }\}$. Moreover, $U_3\in SU(3)$, $U_2\in SU(2)$ implies $e^{6i\theta }=1$, which means that the possible values of $\theta$ are taken from $\left\{\frac{n\pi }{3}:n\in \mathbb{Z}\right\}$, which is discrete.

Question 4. Let $\mathbb{K}=\mathbb{R},\mathbb{C}$ and identify the Lie algebra of an embedded Lie subgroup of $\mathrm{GL}(n,\mathbb{K})$ with a Lie subalgebra of $\mathrm{Mat}(n\times n,\mathbb{K})$. Prove that

•

The Lie bracket on the Lie subalgebra of $\mathrm{Mat}(n\times n,\mathbb{K})$ is the standard commutator of matrices.

•

The Lie algebras of classical groups are the following Lie subalgebras of $\mathrm{Mat}(n\times n,\mathbb{K})$. $$\mathfrak{sl}(n,\mathbb{K})=\{M\in \mathrm{Mat}(n\times n,\mathbb{K}):\mathrm{tr}(M)=0\},$$ $$\mathfrak{o}(n)=\{M\in \mathrm{Mat}(n\times n,\mathbb{K}):M+M^T=0\},$$ $$\mathfrak{u}(n)=\{M\in \mathrm{Mat}(n\times n,\mathbb{K}):M+M^{†}=0\},$$ $$\mathfrak{so}(n)=\mathfrak{o}(n),$$ $$\mathfrak{su}(n)=\{M\in \mathrm{Mat}(n\times n,\mathbb{K}):M+M^{†}=0,\mathrm{tr}(M)=0\}.$$

[Hint : Use the fact that ‘If $H$ is an embedded Lie subgroup of $G$, then $\mathfrak{h}$ is an embedded Lie subalgebra of $\mathfrak{g}$’ and compute the dimension of each matrix group.]

See [Ha17] Theorem 1.5.22 and 1.5.27.

Question 5. Let $X\in \mathfrak{g}$ and $g\in G$ with $(G,\mathfrak{g})$ lying in $$\{(SL(n,\mathbb{K}),\mathfrak{sl}(n,\mathbb{K})),(O(n),\mathfrak{o}(n)),(U(n),\mathfrak{u}(n)),(SO(n),\mathfrak{so}(n)),(SU(n),\mathfrak{su}(n))\}.$$Show that $\mathfrak{g}$ is closed under the map $$X\mapsto g\cdot X\cdot g^{-1}.$$

Solution: Use Question 4 and DIRECT calculation.

Question 6. Let $G$ be a connected Lie group. Define the centre of $G$ as $$Z(G):=\{g\in G:gh=hg,\forall h\in G\}.$$Prove that

•	$Z(G)$ is the kernel of the adjoint representation ${\mathrm{Ad}}_G$.
•	$Z(G)$ is an embedded Lie subgroup in $G$ with Lie algebra given by the centre $\mathfrak{z}(\mathfrak{g})$ of $\mathfrak{g}$.
•	$\mathfrak{g}$ is abelian if and only if $G$ is abelian.
•	${\mathrm{Ad}}_G$ is trivial (i.e. every element of $G$ is mapped to the identity) if and only if $G$ is abelian.
•	The left-invariant and right-invariant vector fields on a connected Lie group $G$ coincide if and only if $G$ is abelian.

Solution: Suppose that ${\mathrm{Ad}}_g$ is trivial, and let $h=\exp (v)\in G$. Now $\gamma (t)=g\cdot \exp \cdot g^{-1}$ is a curve in $G$ whose velocity at $t=0$ is ${\mathrm{Ad}}_g(v)=v$, since ${\mathrm{Ad}}_g$ is trivial. However, $\rho (t)=\exp (tv)$ is also a curve in $G$ with initial velocity $v$, and by the uniqueness of integral curve, $g\cdot \exp (tv)\cdot g^{-1}=\exp (tv)$. In particular, when $t=1$ one has $gh=hg$. For general $h\in G$, since $G$ is connected, by Proposition 7.14 of [Lee13], $h$ can always be written as $h=\exp (v_1)\cdots \exp (v_k)$, and the commutativity comes from iterating what we have done above. Hence $\mathrm{Ker}({\mathrm{Ad}}_g)\subset Z(G)$. On the other hand elements of $Z(G)$ clearly fall into $\mathrm{Ker}({\mathrm{Ad}}_G)$. Therefore $\mathrm{Ker}({\mathrm{Ad}}_G)=Z(G)$.

Now by Proposition 20.8 of [Lee13], we have a commutative diagramSince $\mathrm{Ker}({\mathrm{Ad}}_G)=Z(G)$, the Lie algebra $\mathfrak{z}(\mathfrak{g})$ of $Z(G)$ is given by $\mathrm{Ker}({\mathrm{ad}}_{\mathfrak{g}})=\{X\in \mathfrak{g}:[X,Y]=0,\forall Y\in \mathfrak{g}\}.$ In order this correspondence behaves well, one must demonstrate that $Z(G)$ is a closed subgroup of $G$. For this, let $C_g=\{g_1\in G:g_{1}g=g{g}_1\}$. Then $C_g$ is the preimage $f^{-1}(\{e\})$ where $f$ is the smooth map $f(g_1):=g{g}_1{g}^{-1}{g}_1^{-1}$. Therefore, $C_g$ is closed, and then $Z(G)={\cap }_{g\in G}{C}_g$ is a closed subgroup of $G$.

Reference

[Lee13]	John M. Lee. Introduction to Smooth Manifolds. Graduate Texts in Mathematics. Springer, 2013.
[Ha17]	Hamilton, Mark J.D. Mathematical Gauge Theory. Universitext. Springer, 2017.