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resources:books:baez_muniain [2017/12/27 12:21]
jakobadmin [Solutions to Exercises]
resources:books:baez_muniain [2018/05/05 17:17] (current)
jakobadmin ↷ Page moved from resources:baez_muniain to resources:books:baez_muniain
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 ====== Baez, Muniain: Gauge Fields, Knots and Gravity ====== ====== Baez, Muniain: Gauge Fields, Knots and Gravity ======
 + 
 <tabbox Why is it interesting?> ​ <tabbox Why is it interesting?> ​
 <WRAP imageshadow>​ <WRAP imageshadow>​
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   ​   ​
-<tabbox Reading Notes> ​+<tabbox Reading Notes/Summaries
  
-  * http://​michaelnielsen.org/​blog/​yang_mills.pdf+  * A great summary of Gauge, Fields, Knots and Gravity was [[http://​michaelnielsen.org/​blog/​yang_mills.pdf|published by Michael Nielsen]] ​
  
  
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 <tabbox Solutions to Exercises> ​ <tabbox Solutions to Exercises> ​
  
-  * [[https://​people.phys.ethz.ch/​~renes/​gfkg.html|Solutions by Joseph M. Renes]]. ("​Solutions to nearly all of the exercises in Part I, many of them in Part II, and none in Part III"​) ​ 
-  * https://​ncatlab.org/​brucebartlett/​published/​BaezMunian. (Solutions to all exercises in Part I and Part II.) 
  
  
---> 1# +**Part I:** 
-Q: Let $\vec{k}\in\R^3$ and lt $\omega=|\vec{k}|$. Fix $\vec{E}\in\mathbb{C}^3$ with + 
 +--> ​Exercise ​1# 
 +**Q:** Let $\vec{k}\in\mathbb ​R^3$ and lt $\omega=|\vec{k}|$. Fix $\vec{E}\in\mathbb{C}^3$ with 
 $\vec{k}\cdot\vec{E}=0$ and $\vec{k}\times\vec{E}=i\omega\vec{E}$. Show that  $\vec{k}\cdot\vec{E}=0$ and $\vec{k}\times\vec{E}=i\omega\vec{E}$. Show that 
 $\vec{\mathcal{E}}(t,​\vec{x})=\vec{E}e^{-i(\omega t-\vec{k}\cdot\vec{x})}$ satisfies the $\vec{\mathcal{E}}(t,​\vec{x})=\vec{E}e^{-i(\omega t-\vec{k}\cdot\vec{x})}$ satisfies the
 vacuum Maxwell equations. vacuum Maxwell equations.
  
-A:+**A:**
  
  
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 <-- <--
  
--->2#+-->Exercise ​2#
  
-Q: Given a Lie group $G$, define its identity component $G_0$ to be the connected component ​+**Q:** Given a Lie group $G$, define its identity component $G_0$ to be the connected component ​
 containing the identity element. Show that the identity component of any Lie group is a subgroup, and a Lie containing the identity element. Show that the identity component of any Lie group is a subgroup, and a Lie
 group in its own right. group in its own right.
  
-A: Suppose we have a path from the identity to $g\in G_0$. Now map this + 
 +**A:** Suppose we have a path from the identity to $g\in G_0$. Now map this  
 +path to a new path by multiplying each element by $h\in G_0$. This path starts at $h$ and since the mapping is continuous, must remain in $G_0$. (Otherwise, smoothly mapping the group manifold to $\mathbb R^n$ would show a  
 +discontinuity at some point.) Thus $hg\in G_0$ for all $h,g\in G_0$. There'​s a certain tension between having a smooth manifold with disconnected pieces --- given a map from the  
 +the manifold to itself, one must take care that it does not have a discontinuous action, mapping some points in one component to another component. When this map is an element of the group, this requirement makes 
 +$G_0$ into a subgroup. 
 + 
 +<-- 
 + 
 +**Part II:** 
 + 
 +--> Exercise 1# 
 +**Q:**Given a Lie group $G$, define its identity component $G_0$ to be the connected component  
 +containing the identity element. Show that the identity component of any Lie group is a subgroup, and a Lie 
 +group in its own right. 
 + 
 +**A:​** ​Suppose we have a path from the identity to $g\in G_0$. Now map this 
 path to a new path by multiplying each element by $h\in G_0$. This path starts at $h$ and since the mapping is continuous, must remain in $G_0$. (Otherwise, smoothly mapping the group manifold to $\R^n$ would show a  path to a new path by multiplying each element by $h\in G_0$. This path starts at $h$ and since the mapping is continuous, must remain in $G_0$. (Otherwise, smoothly mapping the group manifold to $\R^n$ would show a 
 discontinuity at some point.) Thus $hg\in G_0$ for all $h,g\in G_0$. discontinuity at some point.) Thus $hg\in G_0$ for all $h,g\in G_0$.
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 the manifold to itself, one must take care that it does not have a discontinuous action, mapping some points in one component to another component. When this map is an element of the group, this requirement makes the manifold to itself, one must take care that it does not have a discontinuous action, mapping some points in one component to another component. When this map is an element of the group, this requirement makes
 $G_0$ into a subgroup. $G_0$ into a subgroup.
- 
 <-- <--
-  ​+ 
 +//Source: [[https://​github.com/​joerenes/​Baez-Muniain-solutions|Solutions to Exercises in Gauge Fields, Knots and Gravity]] by Joseph M. Renes licensed under a Creative Commons Attribution 4.0 Licence.//​ 
 + 
 + 
 +---- 
 + 
 +See also: 
 + 
 +  * [[https://​people.phys.ethz.ch/​~renes/​gfkg.html|Solutions by Joseph M. Renes]]. ("​Solutions to nearly all of the exercises in Part I, many of them in Part II, and none in Part III")  
 +  * https://​ncatlab.org/​brucebartlett/​published/​BaezMunian. (Solutions to all exercises in Part I and Part II.)
 <tabbox Discussions>​ <tabbox Discussions>​
    
resources/books/baez_muniain.1514373687.txt.gz · Last modified: 2017/12/27 11:21 (external edit)