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On page 79, the author states

One is thus concerned only with $\mathbf{Z}$ modulo a component parallel to $\mathbf{V}$, i.e. only with the projection of $\mathbf{Z}$ at each point $q$ into the space $Q_q$ consisting of equivalence classes of vectors which differ only by addition of a multiple of $\mathbf{V}$. This space can be represented as the subspace $H_q$ of $T_q$ consisting of vectors orthogonal to $\mathbf{V}$.

where $\mathbf{Z}$ is the tangent vector $(\partial/\partial t)_\lambda$ on a family $\lambda (t,s)$ of curves, and $\mathbf{V}$ the timelike tangent vector parameterized by $s$.

How do we formally define this quotient space? I'm not very familiar with the concept of equivalence classes, quotient space, isomorphic, modulo and coset; only from a linear algebra course I remembered the professor defined that: given two vector spaces $W \le V$, we define $V/W$ the quotient space of $V$ by $W$ to be the set $V/W=\{ x+W\ |\ x\in V \}$. And something about the square bracket $[x]$ being the coset of $V$ with representative $x$.

On page 86, where the author is talking about the null curves

The second difference is that $Q_q$, the quotient of $T_q$ by $\mathbf{K}$, is not now isomorphic to $H_q$, the subspace of $T_q$ orthogonal to $\mathbf{K}$, since $H_q$ includes the vector $\mathbf{K}$ itself as $g(\mathbf{K},\mathbf{K})=0$. In fact as will be shown below, one is not really interested in the whole of $Q_q$ but only in the subspace $S_q$ consisting of equivalence classes of vectors in $H_q$ which differ only by multiple of $\mathbf{K}$.

where $\mathbf{K}$ is the tangent vector on the null geodesics.

I know that it's not clear how to define a projection of $T_q$ into the subspace $H_q$ orthogonal to $\mathbf{K}$ since $g(\mathbf{K},\mathbf{K})=0$, but I want to understand the concept of quotient space here, and how formally that $Q_q$ is not isomorphic to $H_q$, an from these notions construct a projection operator ${h^a}_b$. What is the subspace $S_q$? These kinds of discussion are all over this chapter 4.

Heuristically (I don't know why) the author used the pseudo-orthonormal bases defined on page 87: take $\mathbf{E}_4$ equal to $\mathbf{K}$ and $\mathbf{E}_3$ some other null vector satisfying $g(\mathbf{E}_3,\mathbf{E}_4)=-1$, and $\mathbf{E}_1$ and $\mathbf{E}_2$ to be unit space like vectors, orthogonal to each other and to $\mathbf{E}_3$ and $\mathbf{E}_4$. The following statements are out of my reach

It can be seen that $\mathbf{E}_1$, $\mathbf{E}_2$ and $\mathbf{E}_4$ constitute a basis for $H_q$ while the projections into $Q_q$ of $\mathbf{E}_1$, $\mathbf{E}_2$ and $\mathbf{E}_3$ form a basis of $Q_q$, and the projections of $\mathbf{E}_1$ and $\mathbf{E}_2$ form a basis of $S_q$. We shall normally not distinguish between a vector $\mathbf{Z}$ and its projection into $Q_q$ or $S_q$. We shall call a basis having the properties of $\mathbf{E}_1$, $\mathbf{E}_2$, $\mathbf{E}_3$, $\mathbf{E}_4$, above, pseudo-orthonormal.

The author further concluded that the dual basis $\mathbf{E}^3$ is equal to $-K^ag_{ab}$ and $\mathbf{E}^4$ is $-L^ag_{ab}$. Well, I think then $L_b = -K^a g_{ab} = -K_b$, and $g(\mathbf{E}_3,\mathbf{E}_4)=0$?

I really like to know these geometrical objects as they are so powerful and abstract, which is good. My current reference book is Schutz, Geometrical methods of mathematical physics. He touches on these idea on a section about cohomology theory (this section is categorized as supplementary topics, and in fact I did not understand a word). Is there any suggestion to a light weighted reference book for a physicist?

Thanks!

raegakj
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  • This is unrelated to cohomology. – Ryan Unger Mar 16 '16 at 14:40
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    A friendly word of advice, OP. If you feel like you're stumbling time and time again in a book, you might not be ready for it. I think Wald's book on GR treats this material more clearly for a beginner. (I will still answer this question when I have time.) – Ryan Unger Mar 16 '16 at 14:47
  • I believe so. In fact, now I'm only interested in the geodesic congruence and I found it was treated in this book, so I just dig right into it.. I think a good mathematical reference book will help. – raegakj Mar 16 '16 at 14:50
  • Actually I've found the lecture notes by Poisson covered the same material in a more detailed manner. But who don't want to read the book by Hawking first? Haha – raegakj Mar 16 '16 at 15:11

1 Answers1

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$H_q$ is indeed a quotient space constructed from $T_q$. We establish the equivalence relation $\mathbf{X}\sim\mathbf{W}$ for $\mathbf{X},\mathbf{W}\in T_q$ if $\mathbf{X}-\mathbf{W}=k\mathbf{V}$ with $k\in\mathbb{R}$. Then $H_q\cong T_q/\sim$. (In "linear algebra notation," we have $T_q/\sim=T_q/\operatorname{span}\mathbf{V}$.$^1$) To see this more clearly, let $[\mathbf{X}]$ be the equivalence class of $\mathbf{X}$ under $\sim$. Expand $\mathbf{X}$ in the basis $\{\mathbf{E}_1,\mathbf{E}_2,\mathbf{E}_3,\mathbf{V}\}$ as $X^a \mathbf{E}_a$. (We set $\mathbf{E}_4=\mathbf V$ for convenience.) Let $\mathbf W\in [\mathbf X]$, then $\mathbf W=W^a\mathbf E_a$ and $$\mathbf X-\mathbf W=(X^4-W^4)\mathbf V+(X^\alpha-W^\alpha)\mathbf E_\alpha=k\mathbf V$$ where $\alpha=1,2,3$. Thus $X^\alpha=W^\alpha$ and $[\mathbf X]$ uniquely determines ${}_\bot \mathbf X:=X^\alpha\mathbf E_\alpha$. As explained here, ${}_\bot \mathbf X\in H_q.$ Likewise, one can show that $\mathbf X\in H_q$ gives rise to an equivalence class $[\mathbf X]$. Thus the quotient space and the orthogonal subspace are isomorphic.

We will show that $Q_q:=T_q/\operatorname{span}\mathbf K\not\cong H_q$. Note that $H_q:=\{\mathbf{X}\in T_q\mid g(\mathbf{X},\mathbf{K})=0\}$. From the above discussion, it should be clear that $\mathbf K\notin Q_q$. But since $\mathbf K$ is null, $\mathbf K\in H_q$. This cannot happen if $H_q\cong Q_q$.

The space $S_q$ is the set of vectors orthogonal to $\mathbf K$ modulo $\mathbf K$ itself, that is, $S_q=H_q/\operatorname{span}\mathbf K$.

To see that $\mathbf E_1,\mathbf E_2$ and $\mathbf K$ span $H_q$, simply note that $H_q$ is $3$-dimensional and $\operatorname{span}\{\mathbf E_1,\mathbf E_2,\mathbf K\}$ contains only vectors orthogonal to $\mathbf K$ and is $3$-dimensional.

The proof that $S_q=\operatorname{span}\{\mathbf E_1,\mathbf E_2\}$ is identical to the proof in the first paragraph and is left to the reader.

The concept of a dual basis is explained on page 17. We explain why $E^3_a=-K^bg_{ba}$ and leave the reader to verify the rest: $\mathbf{E}^3(\mathbf{E}_3)=E^3_aE^a_3=-K^bg_{ba}L^a=-\langle\mathbf K,\mathbf L\rangle=1=\delta^3{}_3$.

$^1$ The span of a single vector $\mathbf X$ is $\{a\mathbf X\mid a\in\mathbb{R}\}$.

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Ryan Unger
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  • For $Q_q:=T_q/ \text{span} \mathbf{V}$, we only show that $[\mathbf{X}]$ is uniquely determined by ${}\bot \mathbf{X}$, where ${}\bot \mathbf{X} \in H_q$. What do you mean by "Likewise, one can show that $\mathbf{X} \in H_q$ gives rise to an equivalence class $[\mathbf{X}]$"? Isn't it a definition? How do we conclude from this that $H_q \cong Q_q$? Maybe there are other elements in $H_q$ that is not in $Q_q$. – raegakj Mar 17 '16 at 03:38
  • So, we construct $S_q$ so that $\text{span}\mathbf{K} \notin S_q$. Are $Q_q$ (from $T_q/\text{span}K$) and $S_q$ isomorphic? And what do you mean by the sentence "is uniquely determined by" in the firs part? – raegakj Mar 17 '16 at 09:16
  • @Henry 1. To show that $H_q\cong Q_q$, show that there exists a map $f:H_q\to Q_q, \mathbf X\mapsto[\mathbf X]$ that is injective and surjective. You should be able to do the careful proof yourself, I've laid most of it out in the post. 2. No, $Q_q$ and $S_q$ are not isomorphic, their dimensions are not equal. 3. I mean that the only information you need for $[\mathbf X]$ is ${}\bot \mathbf X$ and one ${}\bot \mathbf X$ generates exactly one equivalence class. – Ryan Unger Mar 17 '16 at 12:17
  • The elements in $Q_q$ are the equivalent classes $[\mathbf{X}]$. How can we say that $\mathbf{K} \notin Q_q?$. I know that the vectors in $\text{span}\mathbf{K}$ are mapped into zero vector, so...? I think I know the concept, but I cannot state it formally. – raegakj Mar 18 '16 at 09:38
  • By $\langle\mathbf{K},\mathbf{L}\rangle$ I think you mean $g(\mathbf{K},\mathbf{L})$. Or is it a common practice to denote the mapped one-form $\mathbf{K} = g(\mathbf{K})$? – raegakj Mar 18 '16 at 10:00
  • @Henry It's standard to denote the inner product $g(-,-)$ induced on $\Gamma(TM)\times\Gamma(TM)$ by $\langle-,-\rangle$. – Ryan Unger Mar 18 '16 at 14:14
  • @Henry Take the vector space $\mathbb{R}^2=\operatorname{span}{\mathbf i,\mathbf j}$ and quotient it by $\mathbb R$. Try to convince yourself that $\mathbb{R}^2/\mathbb R$ contains only the span of the vector $\mathbf i$ and $\mathbf j\notin \mathbb{R}^2/\mathbb{R}$. – Ryan Unger Mar 18 '16 at 14:26
  • In two dimensional Cartesian system, take the y axis to be the subspace, then $Q_q = { \ldots,[(-1,0)],[(0,0)],[(1,0)],\ldots } $, where $[(1,0)]=[(1,1)]=\ldots$ are the same equivalent class. Now the element in the one dimensional vector space $Q_q$ are the equivalent classes, what do we mean by $\mathbf{i}$ and $\mathbf{j}$? – raegakj Mar 18 '16 at 14:47
  • @Henry $\mathbf i=(1,0),\mathbf j=(0,1)$. – Ryan Unger Mar 18 '16 at 17:43
  • Yes, they are these two dimensional element. We cannot compare them without a map to the one dimensional quotient space, right? Maybe there were some description of mine above which were wrong. – raegakj Mar 18 '16 at 17:50
  • @Henry Please see the Wiki article on the quotient of a vector space. The necessary theorem is: if $V=U\oplus W$, then $V/W\cong U$. If you don't understand this, please ask a question on Math SE. – Ryan Unger Mar 18 '16 at 18:21
  • Maybe it is the map $f:H_q \rightarrow Q_q$ is only surjective but non-injective, since both zero vector and $\mathbf{K}$ are mapped to the same equivalent classes $[0]$. This is cool stuff, Thanks a lot!!! Your explanation is always so clear and insightful. I will take an algebra course next semester for sure. – raegakj Mar 19 '16 at 02:02
  • @Henry Uh, note that in the first paragraph, where I'm talking about $f$, $\mathbf K$ has not been introduced yet. I think you're getting things mixed up here. – Ryan Unger Mar 19 '16 at 02:05
  • I don't get it. By $f$ I mean $f:H_q \rightarrow T_q/\text{span}\mathbf{K}$. And it is not because $\mathbf{K} \notin T_q/\text{span}\mathbf{K}$ that make it not isomorphism, but the map. – raegakj Mar 19 '16 at 02:14
  • @Henry I never claimed that was an isomorphism. (In fact, quite the opposite.) – Ryan Unger Mar 19 '16 at 02:15