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I'm reading MTW's Gravitation. On page 92 we have the following statement

$$\langle\omega^{i_1}\wedge\dots\wedge\omega^{i_p},e_{j_1}\wedge\dots\wedge e_{j_p}\rangle = \delta^{i_1\dots i_p}_{j_1\dots j_p}\tag{1}$$

Where $\{e_i\}$ and $\{\omega^{i}\}$ are bases of $V$ and $V^{*}$ satisfying $\omega^{i}(e_{j}) = \delta^{i}_{j}$

However, using the definition of the wedge product $$(\omega\wedge\eta)_{a_1\dots a_{p+q}} = \frac{(p+q)!}{p!q!}(\omega\otimes\eta)_{[a_1\dots a_{p+q}]}$$ I get $$\omega^{i_1}\wedge\dots\wedge\omega^{i_p} = p!\,\omega^{[i_1}\otimes\dots\otimes\omega^{i_p]}$$ $$e_{j_1}\wedge\dots\wedge e_{j_p} = p! \, e_{[j_1}\otimes\dots\otimes e_{j_p]}$$ which gives me $$\langle\omega^{i_1}\wedge\dots\wedge\omega^{i_p},e_{j_1}\wedge\dots\wedge e_{j_p}\rangle\\ = p!p!\langle\omega^{[i_1}\otimes\dots\otimes\omega^{i_p]},e_{[j_1}\otimes\dots\otimes e_{j_p]}\rangle\\ = p!p!\langle\omega^{[i_1},e_{[j_1}\rangle\cdots\langle\omega^{i_p]},e_{j_p]}\rangle\\ = p!p!\delta^{[i_1}_{[j_1}\cdots\delta^{i_p]}_{j_p]}\\ = p!\delta^{i_1\dots i_p}_{j_1\dots j_p}$$

Where did I go wrong? is this a mistake in the book? or is the contraction on the left hand side of $(1)$ defined to be equal to $\delta^{i_1\dots i_p}_{j_1\dots j_p}$?

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  • $\begingroup$ Is there a combinatorial factor included in the antisymmetrization of indices? I do not have MTW at home, it's a bit much to carry around. Sometimes, the $A_{[i,j]}$ means $\frac{1}{2\!}(A_{ij}-A_{ji})$. In that nomenclature, your wedge products give in the case $p=2$, $\omega^1 \wedge \omega^2 = \omega^1 \otimes \omega^2-\omega^2 \otimes \omega^1$. Many texts use this convention for the wedge product of two one-forms. $\endgroup$ – James S. Cook Jul 10 '18 at 2:52
  • $\begingroup$ That's right. The antisymmetrization brackets over $k$ indices include the factor $\frac{1}{k!}$. That's why there is a factor of $(p+q)!$ in my definition of the wedge. And is using these conventions that I got the problem, since this is the convention used in MTW. $\endgroup$ – Jackozee Hakkiuz Jul 10 '18 at 2:57
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    $\begingroup$ @pglpm yeah, I know. The problem is that geometers and algebraists use slightly different conventions to embed the exterior algebra into the tensor algebra, so one needs to check everytime what conventions are used in which book. The problem here is that it looked to me like two different conventions were used in the same book. Frankly, I posted this a while ago, and I need to take a look at my notes again to see what exactly is the problem. After that I'll come back and edit the question adding a more detailed exposition of the apparent problem. Thanks for reviving the post, by the way. :) $\endgroup$ – Jackozee Hakkiuz Apr 28 at 22:35
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    $\begingroup$ I checked Schouten, and from its formula (7.5), § II.7, p. 28, it seems that contraction of multi-forms and multi-vectors does include an extra factor $1/q!$ with respect to the usual tensor contraction. $\endgroup$ – pglpm Apr 29 at 7:30
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    $\begingroup$ @pglpm That looks right. Would you mind writing it as an answer below so that I can accept it? $\endgroup$ – Jackozee Hakkiuz Apr 29 at 8:23
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The formula in Misner & al (Box 4.1, p. 92, item 4.) is correct, but they could have been clearer in explaining its meaning.

The notation "$\langle\;,\;\rangle$" in Gravitation denotes the dual product (also called "inner product" by some authors) of a vector space and its dual. In this case the vector space is the exterior space $\Lambda^p(V)$ of $p$-vectors based on a vector space $V$ of dimension $n$, and the dual is the exterior space of $p$-covectors $\Lambda^p(V^*)$.

In coordinate-free form, this dual product is defined as follows:

  1. For a covector $\alpha$ and a vector $a$ it's defined in the usual way;

  2. For a simple $p$-covector (that is, one that can be written as the exterior product of $p$ covectors) $\eta = \eta^1 \wedge \dotsb \wedge \eta^p$, and a simple $p$-vector $w = w_1 \wedge \dotsb \wedge w_p$, it's defined as $$\langle \eta, w \rangle := \eta(w_1, \dotsc, w_p) \equiv \sum_{\sigma} \mathrm{sgn}(\sigma)\; \langle\eta^{\sigma_1}, w_1\rangle \dotsm \langle\eta^{\sigma_p}, w_p\rangle,$$ where the sum is over every permutation $\sigma$ of $\{1,\dotsc,p\}$ with sign $\mathrm{sgn}(\sigma)$. It can be shown that this definition doesn't depend on the specific representation of the simple $p$-(co)vector in terms of exterior product of $1$-(co)vectors. Note: simple (co)vectors are also called "monomials" (Choquet-Bruhat & al) or "extensors" (Barnabei & al).

  3. For non-simple $p$-covectors and $p$-vectors it's extended by bilinearity.

From this definition we also obtain the coordinate representation of the dual product. Given a basis $(e_1, \dotsc, e_n)$ for $V$ and dual basis $(\omega^1, \dotsc, \omega^n)$, we build an associated basis $(e_{J_1} \wedge\dotsb\wedge e_{J_p})$ for $\Lambda^p(V)$ and the dual $(\omega^{I_1} \wedge\dotsb\wedge \omega^{I_p})$ for $\Lambda^p(V^*)$. I'm using Choquet-Bruhat & al's convention: capitalized indices $J_k$ mean $$J_1 < \dotso < J_p$$ (Misner, Thorne, Wheeler use the convention with indices within bars "$\lvert\;\rvert$" instead). This restriction on the indices is very important, because the dimension of $\Lambda^p(V)$ is $\binom{n}{p}$, different from the dimension of the tensor space $\otimes^p V$ in which it could be embedded. Compare the remarks in Choquet-Bruhat & al, p. 197, about "strict components".

These bases are clearly dual from the definition 2. above. Then, by 3., in coordinates the dual product of a $p$-vector with components $A^{J_1\dotso J_p}$ and a $p$-covector $\alpha_{I_1 \dotso I_p}$ is simply \begin{equation}A^{J_1\dotso J_p}\;\alpha_{J_1 \dotso J_p}\label{dual}\tag{1} \end{equation} with Einstein's summation convention.

This is also the expression given in Gravitation.

The crucial step is point 2. above: the dual product does not correspond to a contraction of the antisymmetric tensors corresponding to the $p$-covector and $p$-vector. Formally there's a $p!$ factor difference; but the difference is deeper, because we're operating on different spaces. The exterior spaces $\Lambda^p(V)$ and $\Lambda^p(V^*)$ don't have a tensor product. As Deschamps, § III.4.1, p. 119 says, "The exterior algebra can be introduced independently of tensors and this result is some economy of notation".

Some extra info

The dual product above can be generalized to a product between a $p$-covector $\alpha$ and a $q$-vector $a$; I denote it here with $\alpha\vert a$. If $p > q$ the result is a covector, and if $p < q$, a vector. Lindell (§ 1.3.4) calls this the "incomplete dual product"; Deschamps (§ 4.6) calls this just "inner product". Its coordinate-free definition for $p > q$ is this: $\alpha\vert a$ is the $(p-q)$-covector such that $$\langle\alpha\vert a,\; x\rangle = \langle \alpha,\; a\wedge x\rangle \quad\text{for every $(p-q)$-vector $x$}.$$

Its coordinate expression is $$ (\alpha\vert a)_{I_1\dotso I_{p-q}} = \alpha_{K_1\dotso K_p}\; \epsilon_{J_1\dotso J_q\;I_1\dotso I_{p-q}}^{K_1 \dotso K_p}\; a^{J_1\dotso J_q}\;x^{I_1\dotso I_{p-q}}. $$ It can be found combining the expression $\ref{dual}$ for the dual and the expression for the exterior product, given for example by Choquet-Bruhat & al, p. 198 (bottom): $$(a\wedge x)^{K_1 \dotso K_p} = \epsilon_{J_1\dotso J_q\;I_1\dotso I_{p-q}}^{K_1 \dotso K_p}\; a^{J_1\dotso J_q}\;x^{I_1\dotso I_{p-q}}.$$ Here $\epsilon$ is the Kronecker tensor (see for example Choquet-Bruhat & al, p. 142).

This product generalizes the dual product discussed above and also the interior product of a $p$-covector $\alpha$ and a $1$-vector $v$, often denoted $\mathrm{i}_v\alpha$. It is often used with the volume element, to transform a $q$-vector into an $(n-q)$-covector (and vice versa with the inverse volume element): see for example Schouten eq. (7.5), § II.7, p. 28. This product also appears, implicitly, in the operation of meet defined in Peano spaces: the meet of two multi-vectors is actually the generalized dual product of both with a volume element; see for example Barnabei & al p. 132, or White p. 98.

This product looks like a contraction, but it has many important differences from the usual contraction in a tensor space. In a tensor space, contraction requires the specification of indices (or argument slots); in an exterior space this specification doesn't matter (owing to the antisymmetrization).

Finally, the fact that $\langle \omega^{I_1}\wedge\dotsb\wedge\omega^{I_p}, e_{J_1}\wedge\dotsb\wedge e_{J_p}\rangle = \delta^{I_1\dotso I_p}_{J_1\dotso J_p}$– for example $\langle\mathrm{d}x \wedge\mathrm{d}y, \partial_x \wedge \partial_y\rangle =1$ – makes also sense geometrically: the 2-covector represents a tube with an outer orientation, and the 2-vector a planar area with an inner orientation, and their product tells us how many times the cross-sectional area of the tube fills the planar area (a notion which only requires affine geometry, that is, parallelism). With the dual bases these areas are the same and the result is unity.

The generalized inner product extends this geometric meaning, as beautifully illustrated by Burke (1995, for example p. 17.4) and Schouten (chap. II).

References:

  • M. Barnabei, A. Brini, G.-C. Rota: On the exterior calculus of invariant theory. J. Algebra 96/1 (1985), 120–160.

  • W. L. Burke: Div, Grad, Curl Are Dead (1995) http://people.ucsc.edu/~rmont/papers/Burke_DivGradCurl.pdf.

  • Y. Choquet-Bruhat, C. DeWitt-Morette, M. Dillard-Bleick: Analysis, Manifolds and Physics. Part I: Basics (rev. ed. Elsevier 1996).

  • G. A. Deschamps, E. M. de Jager, F. John, J. L. Lions, N. Moisseev, F. Sommer, A. N. Tihonov, V. Tikhomirov, A. B. Vasil'eva, V. M. Volosov, D. J. A. Welsh, T. Yamanouchi: Mathematics applied to physics (Springer 1970).

  • I. V. Lindell: Differential Forms in Electromagnetics (IEEE & Wiley 2004).

  • C. W. Misner, K. S. Thorne, and J. A. Wheeler: Gravitation (25th printing, Freeman 2003).

  • J. A. Schouten: Tensor Analysis for Physicists (2nd ed. Dover 1989).

  • N. L. White (ed.): Invariant Methods in Discrete and Computational Geometry (Springer 1995).

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    $\begingroup$ Actually, your answer helped me clear some questions I had but didn't take the time to formulate. :) Thanks a lot. $\endgroup$ – Jackozee Hakkiuz Apr 30 at 22:23

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