# MTW the contraction of a basis bi-vector with a basis two-form. Why am I getting this factor of 2?

My question pertains to the discussion of two-forms and bi-vectors in MTW, Chapters 3 and 4. I set out to understand how the expression for the contraction of a basis p-vector with a basis p-form results form the basic definitions in the case of bi-vectors and two-forms. I end up with a factor of 2 which don't understand.

That is, starting from the definitions of coordinate basis vectors, dual basis one-forms, the tensor product and thereby the wedge product, I obtain

$$\left\langle \mathfrak{e}_{\alpha}\wedge\mathfrak{e}_{\beta},\mathbf{\omega}^{\mu}\wedge\mathbf{\omega}^{\nu}\right\rangle =2\delta_{\alpha\beta}^{\mu\nu}.$$

The result indicated in Box 4.1 A-4 is

$$\left\langle \mathfrak{e}_{\alpha}\wedge\mathfrak{e}_{\beta},\mathbf{\omega}^{\mu}\wedge\mathbf{\omega}^{\nu}\right\rangle =\delta_{\alpha\beta}^{\mu\nu}.$$

In the referenced Exercise 4.12 the result appears to be given by definition. But that definition should be consistent with results arising from other definitions.

My question is: why am I getting a factor of two when working from the basic definition?

One point of uncertainty is what it means to contract the tensor product of two basis vectors with the tensor product of two basis one-forms. Since the wedge product is defined as a difference of the tensor products, I contract the wedge products using the tensor product forms according what seems correct to me.

Here's my development: The components of the covariant basis vectors of a coordinate basis with respect to itself are simply the elements of the columns of the identity matrix. So

$$\mathfrak{e}_{\alpha}=\mathfrak{e}_{\sigma}\delta_{\alpha}^{\sigma}.$$

Similarly, the components of corresponding contravariant basis one-forms are rows of the identity matrix

$$\mathbf{\omega}^{\alpha}\equiv\mathbf{d}x^{\alpha}=\delta_{\sigma}^{\alpha}\mathbf{\omega}^{\sigma}.$$

The contractions of the basis vectors with the basis one-forms are

$$\left\langle \mathfrak{e}_{\alpha},\mathbf{\omega}^{\beta}\right\rangle =\delta_{\alpha}^{\sigma}\delta_{\sigma}^{\beta}=\delta_{\alpha}^{\beta}.$$

Following the discussion of Exercise 3.4, I conclude that we may write the tensor product of two basis vectors as

$$\mathfrak{T}_{\alpha\beta}=\mathfrak{e}_{\alpha}\otimes\mathfrak{e}_{\beta}=\mathfrak{e}_{\sigma}\otimes\mathfrak{e}_{\tau}\delta_{\alpha}^{\sigma}\delta_{\beta}^{\tau}=\mathfrak{e}_{\sigma}\otimes\mathfrak{e}_{\tau}T_{\alpha\beta}^{\sigma\tau}.$$

Here each of the$$\mathfrak{T}_{00},\mathfrak{T}_{01},\mathfrak{T}_{23},$$ etc., is an individual tensor. Thus the components of $$\mathfrak{T}_{23},$$ and, in general $$\mathfrak{T}_{\alpha\beta}$$ are

$$\mathfrak{T}_{23}=\left\{ T_{23}^{\sigma\tau}\right\} =\left\{ \delta_{2}^{\sigma}\delta_{3}^{\tau}\right\} \text{ and }\mathfrak{T}_{\alpha\beta}=\left\{ T_{\alpha\beta}^{\sigma\tau}\right\} =\left\{ \delta_{\alpha}^{\sigma}\delta_{\beta}^{\tau}\right\} .$$

Similarly the tensor products of the basis one-forms are

$$\mathbf{\omega}^{\mu}\otimes\mathbf{\omega}^{\nu}=\mathbf{\Omega}^{\mu\nu}=\left\{ W_{\sigma\tau}^{\mu\nu}\right\} =\left\{ \delta_{\sigma}^{\mu}\delta_{\tau}^{\nu}\right\} .$$

Now, the contractions of the $$\mathfrak{T}_{\alpha\beta}$$ with the $$\mathbf{\Omega}^{\mu\nu}$$ will be

$$\left\langle \mathfrak{T}_{\alpha\beta},\mathbf{\Omega}^{\mu\nu}\right\rangle =T_{\alpha\beta}^{\sigma\tau}W_{\sigma\tau}^{\mu\nu}=\delta_{\alpha}^{\sigma}\delta_{\beta}^{\tau}\delta_{\sigma}^{\mu}\delta_{\tau}^{\nu}=\delta_{\alpha}^{\mu}\delta_{\beta}^{\nu}.$$

Following the pattern of Equation 4.2 the wedge products of basis vectors (producing basis bi-vectors) and basis one forms (producing basis two-forms) are

$$\mathfrak{e}_{\alpha}\wedge\mathfrak{e}_{\beta}\equiv\mathfrak{e}_{\alpha}\otimes\mathfrak{e}_{\beta}-\mathfrak{e}_{\beta}\otimes\mathfrak{e}_{\alpha}$$

$$=\left\{ T_{\alpha\beta}^{\sigma\tau}-T_{\beta\alpha}^{\sigma\tau}\right\} =\left\{ \delta_{\alpha}^{\sigma}\delta_{\beta}^{\tau}-\delta_{\beta}^{\sigma}\delta_{\alpha}^{\tau}\right\} \equiv\left\{ \delta_{\alpha\beta}^{\sigma\tau}\right\} ,$$

and

$$\mathbf{\omega}^{\mu}\wedge\mathbf{\omega}^{\nu}\equiv\mathbf{\omega}^{\mu}\otimes\mathbf{\omega}^{\nu}-\mathbf{\omega}^{\nu}\otimes\mathbf{\omega}^{\mu}$$

$$=\left\{ W_{\sigma\tau}^{\mu\nu}-W_{\sigma\tau}^{\nu\mu}\right\} =\left\{ \delta_{\sigma\tau}^{\mu\nu}\right\} .$$

Now we contract the basis bi-vectors with the basis two-forms

$$\left\langle \mathfrak{e}_{\alpha}\wedge\mathfrak{e}_{\beta},\mathbf{\omega}^{\mu}\wedge\mathbf{\omega}^{\nu}\right\rangle =\delta_{\alpha\beta}^{\sigma\tau}\delta_{\sigma\tau}^{\mu\nu}$$

$$=\left(T_{\alpha\beta}^{\sigma\tau}-T_{\beta\alpha}^{\sigma\tau}\right)\left(W_{\sigma\tau}^{\mu\nu}-W_{\sigma\tau}^{\nu\mu}\right)$$

$$=\left(T_{\alpha\beta}^{\sigma\tau}W_{\sigma\tau}^{\mu\nu}+T_{\beta\alpha}^{\sigma\tau}W_{\sigma\tau}^{\nu\mu}\right)-\left(T_{\alpha\beta}^{\sigma\tau}W_{\sigma\tau}^{\nu\mu}+T_{\beta\alpha}^{\sigma\tau}W_{\sigma\tau}^{\mu\nu}\right)$$

$$=\left(\delta_{\alpha}^{\mu}\delta_{\beta}^{\nu}+\delta_{\beta}^{\nu}\delta_{\alpha}^{\mu}\right)-\left(\delta_{\alpha}^{\nu}\delta_{\beta}^{\mu}+\delta_{\beta}^{\mu}\delta_{\alpha}^{\nu}\right)=2\delta_{\alpha\beta}^{\mu\nu}.$$

As another example, suppose we have two real-valued functions $$f^1$$ and $$f^2$$ on a 2-dimensional manifold. Following Box 4.1 A.4.b,

$$\left\langle \mathbf{d}f^{1}\wedge\mathbf{d}f^{2},\frac{\partial\mathscr{P}}{\partial x^{1}}\wedge\frac{\partial\mathscr{P}}{\partial x^{2}}\right\rangle =\left|\begin{bmatrix}\frac{\partial f^{1}}{\partial x^{1}} & \frac{\partial f^{1}}{\partial x^{2}}\\ \frac{\partial f^{2}}{\partial x^{1}} & \frac{\partial f^{2}}{\partial x^{2}} \end{bmatrix}\right|.$$

Writing this out using the definition of the wedge product in terms of the tensor product, and using juxtaposition to indicate the tensor product, gives: $$\left\langle \mathbf{d}f^{1}\otimes\mathbf{d}f^{2}-\mathbf{d}f^{2}\otimes\mathbf{d}f^{1},\frac{\partial\mathscr{P}}{\partial x^{1}}\otimes\frac{\partial\mathscr{P}}{\partial x^{2}}-\frac{\partial\mathscr{P}}{\partial x^{2}}\otimes\frac{\partial\mathscr{P}}{\partial x^{1}}\right\rangle$$

$$=\left(\left\langle \mathbf{d}f^{1}\mathbf{d}f^{2},\mathfrak{e}_{1}\mathfrak{e}_{2}\right\rangle +\left\langle \mathbf{d}f^{2}\mathbf{d}f^{1},\mathfrak{e}_{2}\mathfrak{e}_{1}\right\rangle \right)-\left(\left\langle \mathbf{d}f^{1}\mathbf{d}f^{2},\mathfrak{e}_{2}\mathfrak{e}_{1}\right\rangle +\left\langle \mathbf{d}f^{2}\mathbf{d}f^{1},\mathfrak{e}_{1}\mathfrak{e}_{2}\right\rangle \right)$$

$$=2\left(\frac{\partial f^{1}}{\partial x^{1}}\frac{\partial f^{2}}{\partial x^{2}}-\frac{\partial f^{1}}{\partial x^{2}}\frac{\partial f^{1}}{\partial x^{2}}\right)=2\left|\begin{bmatrix}\frac{\partial f^{1}}{\partial x^{1}} & \frac{\partial f^{1}}{\partial x^{2}}\\ \frac{\partial f^{2}}{\partial x^{1}} & \frac{\partial f^{2}}{\partial x^{2}} \end{bmatrix}\right|.$$

If we write the contraction of a two-form with the tensor product of the basis vectors, we get the determinant without a factor of 2.

$$\left\langle \mathbf{d}f^{1}\wedge\mathbf{d}f^{2},\frac{\partial\mathscr{P}}{\partial x^{1}}\otimes\frac{\partial\mathscr{P}}{\partial x^{2}}\right\rangle$$

$$=\left\langle \mathbf{d}f^{1}\mathbf{d}f^{2}-\mathbf{d}f^{2}\mathbf{d}f^{1},\frac{\partial\mathscr{P}}{\partial x^{1}}\frac{\partial\mathscr{P}}{\partial x^{2}}\right\rangle =\left|\begin{bmatrix}\frac{\partial f^{1}}{\partial x^{1}} & \frac{\partial f^{1}}{\partial x^{2}}\\ \frac{\partial f^{2}}{\partial x^{1}} & \frac{\partial f^{2}}{\partial x^{2}} \end{bmatrix}\right|$$

The last example is how I would do this following Edwards's Advance Calculus of Several Variables. I have found very few errors in MTW, so it is difficult for me to believe they have this wrong, but it sure looks wrong to me.

My argument boils down to the following:

$$\left(\mathbf{\omega}^{\alpha_{2}}\wedge\mathbf{\omega}^{\alpha_{2}}\wedge\mathbf{\omega}^{\alpha_{3}}\right)_{\beta_{1}\beta_{2}\beta_{3}}=\delta_{\beta_{1}\beta_{2}\beta_{3}}^{\alpha_{1}\alpha_{2}\alpha_{3}}$$

$$\left(\mathfrak{e}_{\gamma_{1}}\wedge\mathfrak{e}_{\gamma_{2}}\wedge\mathfrak{e}_{\gamma_{3}}\right)^{\beta_{1}\beta_{2}\beta_{3}}=\delta_{\gamma_{1}\gamma_{2}\gamma_{3}}^{\beta_{1}\beta_{2}\beta_{3}}$$

$$\left\langle \mathbf{\omega}^{\alpha_{2}}\wedge\mathbf{\omega}^{\alpha_{2}}\wedge\mathbf{\omega}^{\alpha_{3}},\mathfrak{e}_{\gamma_{1}}\wedge\mathfrak{e}_{\gamma_{2}}\wedge\mathfrak{e}_{\gamma_{3}}\right\rangle$$

$$=\delta_{\beta_{1}\beta_{2}\beta_{3}}^{\alpha_{1}\alpha_{2}\alpha_{3}}\delta_{\gamma_{1}\gamma_{2}\gamma_{3}}^{\beta_{1}\beta_{2}\beta_{3}}=\frac{1}{3!}\delta_{\gamma_{1}\gamma_{2}\gamma_{3}}^{\alpha_{1}\alpha_{2}\alpha_{3}}.$$

Which is not the result advertised in Box 4.1. It does, however, follow from definitions given in Exercise 4.12. I am out of time for today, so I will have to explain things more clearly when I get a chance.

• I'm going with "the book is wrong". I will explain in a posted answer. In the meantime, if anyone cares to preempt my folly with an answer showing the error of my ways, I would be grateful. My argument will be that the definitions given in Exercise 4.12 do not lead to the results stated in Box 4.1 and cited above. There is a missing $!p$. – Steven Thomas Hatton Mar 3 at 1:34
• If no-one can provide a satisfactory answer, Steven, you might be a bit cheeky and try emailing Kip Thorne of MTW himself? There's a 2017 email address on iau.org/administration/membership/individual/3565 ; - > – iSeeker Mar 3 at 19:05
• I've thought about it. – Steven Thomas Hatton Mar 4 at 1:47
• I recommend you edit the question and delete everything below "My question is:..." show us how you got the extra factor of $2$ in the first equation. Everything below your question is a distraction. Note, the inner product is in the sense that the $p$-vector is the dual to the $p$-covector. If you're using the tensor definition of the wedge product you'll get the wrong answer. The correct answer is $\left\langle \mathfrak{e}_{\alpha}\wedge\mathfrak{e}_{\beta},\mathbf{\omega}^{\mu}\wedge\mathbf{\omega}^{\nu}\right\rangle =\delta_{\alpha\beta}^{\mu\nu}$ as stated in MTW. – Cinaed Simson Mar 4 at 7:09
• P-forms are completely antisymmetric covariant tensors. P-multivectors are completely antisymmetric contravariant tensors. The contraction of one with the other is tensor contraction. The last equation in my question is the result of carefully following all the definitions given in exercise 4.12, which is the reference given for the result that I claim is wrong. – Steven Thomas Hatton Mar 4 at 7:35

In one way or the other, the book is wrong. The offending equation is the "definition" indicated by the $$\equiv$$ in

.

To my knowledge that indicated equivalence between the alternating generalized Kronecker delta and the contraction of a basis p-form with a basis p-vector is not explained elsewhere in the book. If we take it to be the definition of how to perform such a contraction, then we have the problem that equivalent statements should be interchangeable in mathematical expressions, while these are not. Specifically, as demonstrated in my original question, the result of contraction using the definition of the wedge product in terms of the tensor product produces a result differing by a factor of $$p!.$$

If we take it to be the result of applying the rules of tensor contraction, then it is simply an error of derivation.

Apparently the authors intended the definition of the contraction to give the result shown in #4. above. This is shown in the following example from the book, which is also discussed in my original question:

Based on the comments of Cinaed Simson and Chapter 10 of Electricity and Magnetism for Mathematicians, by Thomas A. Garrity, I now understand why MTW use that definition. But I stand by the assertion that it contradicts the previously given definition, and without a clear demarkation of context (which MTW do not provide) the contradiction amounts to an error. I suspect this is common usage, but has the fault I have demonstrated. That is, one can always express a k-form in terms of the tensor product, and thus arrive at a different result.

Since the inner product of k-forms involves pairs of elements of the vector space spanned by the k-form basis, I suggest writing $$\left\langle \omega,\rho\right\rangle _{k}$$ to distinguish it from the previously defined "contraction". The same will apply to the "contraction" of a k-vector with a k-form.