# Definition of cross product in Spivak's 'Calculus on Manifolds'

1. Aug 11, 2007

### quasar987

1. The problem statement, all variables and given/known data
In Calculus on Manifold pp.83-84, Spivak writes that "if v_1,...,v_{n-1} are vectors in R^n and f:R^n-->R is defined by f(w)=det(v_1,...,v_{n-1},w), then f is an alternating 1-tensor on R^n; therefore there is a unique z in R^n such that <w,z>=f(w) (and this z is denoted v_1 x ... x v_{n-1} and called the cross product of v_1,...,v_{n-1})."

I do not get how he makes that conclusion about the existence and uniqueness of z.

Existence is clear as such a z is easy to construct explicitly. But surely, z is not unique! Take for instance n=3, in which case <w,z>=|w||z|cosO, so that given a lenght |z| for z, it suffices to give it the angle O = arccos(f(w)/[|w||z|]) for the equality <w,z>=f(w) to hold, and evidently, there are an infinity of possible such lenght-angle combinations.

2. Aug 11, 2007

### quasar987

a precise counter-example would be if the v_i are linearly dependant. In this case f(w) = 0 for any w. We then can arrange for <w,z> to vanish by letting z=0 or z=(1/w_1, -1/w_2, 0,...,0).

3. Aug 11, 2007

### AlephZero

You are given the vectors v_1 ... v_{n-1}.

You then define a function of w, using v1 ... v_{n-1}, for ANY vector w.

The uniqueness part says there is a unique vector z that satisfies the condition for ALL possible vectors w.

In your "precise counter example", that unique vector is z = 0. You can't choose a different vector z for every different vector w.

All this is consistent with the "elementary" definition of a vector product in 3-space:

v1 x v2 = 0 when v1 and v2 are parallel vectors
(v1xv2).v3 is the volume of the parallepiped defined by v1,v2, and v3, which can be written as a det(v1,v2,v3).

4. Aug 11, 2007

### quasar987

It makes more sense already! :tongue:

5. Dec 24, 2009

### cmendnba

This follows from a theorem in linear algebra. Let $$V$$ be a real inner product space with inner product $$\left\langle,\right\rangle$$ For a fixed $$\beta\in V$$, the map $$f_{\beta}:V\rightarrow\Re$$ (where $$\Re$$ is the set of real numbers) defined by $$f_{\beta}\left(w\right)=\left\langle w,\beta \right\rangle$$ is easily seen to be a linear functional. It is not difficult to see that every linear functional arises in this manner for a unique choice of $$\beta$$. For a proof, see Hoffman and Kunze's Linear Algebra Ch. 8.3. Also, there is no reason to say alternating $$1$$ tensor because a $$1$$ tensor is just a linear functional and is vacuously alternating (or simply observe that the dimension of the space of alternating $$1$$ tensors is the same as the space of all $$1$$ tensors since $$k=1$$ so that the two spaces are the same). Therefore, all that needs to be noted from Spivak's comments is that the map $$\varphi$$ that he defined is a linear functional. Hence, the linear algebra theorem applies. This doesn't follow from anything that Spivak gives you in the text. The author expects that you have a savvy knowledge of Linear Algebra and Calculus (from a mathematician's pov) before attempting this book.

Last edited: Dec 24, 2009
6. Dec 25, 2009

### AEM

I am curious. What is the relationship between the cross product defined above and the cross product of two vectors, a and b defined by:

(1) a X b is perpendicular to the plane defined by a and b,
(2) a X b has magnitude ab sin(theta),
(3) the vectors a, b, and a X b form a right-handed set.

(For typing simplicity I have not used bold type notation and the a and b coefficients to sin(theta) are the magnitudes of the vectors.)

Quoting Doran and Lasenby in "Geometric Algebra for Physicists", "The cross product has one major failing -it only exists in three dimensions." The primary reason is that in four and higher dimensions the concept of a vector orthogonal to a pair of vectors is not unique. This lack of a cross product in other than three dimensions is often cited as an advantage of Geometric Algebra over traditional vector analysis.
However, if I read the initial post above correctly, it implies something Spivak calls a cross product that exists in R^n, with no restriction on n. Please explain.

7. Dec 25, 2009

### cmendnba

The cross product defined above is the cross product. The cross product of $$n-1$$ vectors in $$\Re^{n}$$ is always defined in this way, but it depends on an a priori choice of $$n-1$$ vectors. If $$n=3$$, then this definition gives a rule for the cross product of two vectors in $$\Re^{3}$$ which is the product you are referring to that one learns in elementary calculus. In this case all these properties you have listed can be proved as theorems.

8. Dec 26, 2009

### AEM

Could you perhaps give me an example in $$\Re^{4}$$ or $$\Re^{5}$$so that I could see how this works?

9. Dec 26, 2009

### D H

Staff Emeritus
This last item necessitates a three dimensional space. If you are looking for a generalization, this has to be dropped. If you are willing to relax the (implied) condition that the cross product is a binary operator one generalization of the cross product is the product of n-1 n-vectors as described in the original post. Suppose you define the cross product as a binary operator $\times$ that is
• Linear, $(a\mathbf x + b\mathbf y)\times \mathbf z = a\mathbf x\times \mathbf z + b\mathbf y\times \mathbf z$ and similarly for the right-hand side of the product
• Orthogonal, $\mathbf x \cdot (\mathbf x \times \mathbf y) = \mathbf y \cdot (\mathbf x \times \mathbf y) = 0$
• Geometrical, $||\mathbf x \times \mathbf y||^2 = ||\mathbf x||^2||\mathbf y||^2 - (\mathbf x \cdot \mathbf y)^2$, or $||\mathbf x \times \mathbf y|| = ||\mathbf x||\,||\mathbf y||\, |\sin\theta|$

The above is almost unique to three dimensions. There is a seven dimensional cross product that satisfies all of the above. Yet another way to define the cross product is in terms of a commutator. The 3-vectors as we now know them were developed from the quaternions. Think of a quaternion as a scalar real part and an imaginary vectorial part. The symmetric difference $(xy-yz)/2$ of two quaternions is a pure imaginary quaternion whose imaginary part is equal to the vector cross product of their imaginary parts. This extends to the octonions, and hence to a seven dimensional cross product. It does not extend any further than that because each the next step up the Cayley–Dickson algebras hierarchy, the sedenions, are not alternative.

10. Dec 26, 2009

### cmendnba

Sure. Once $$n\geq 5$$, the computations get too tedious so we'll do one in $$\Re^{4}$$. Say $$v_{1}=\left(0,1,0,1\right), v_{2}=\left(1,0,2,1\right),$$ and $$v_{3}=\left(2,1,1,0\right)$$. We will compute $$v_{1}\times v_{2}\times v_{3}$$. Define $$\varphi:\Re^{n}\rightarrow \Re$$ by $$\varphi\left(w\right)=det\left(v_{1},v_{2},v_{3},w\right)$$. Write $$w=\left(w_{1},w_{2},w_{3},w_{4}\right)$$. Then you can compute (the computation is tedious) $$\varphi\left(w\right)=det\left(v_{1},v_{2},v_{3},w\right)=w_{1}-3w_{2}+3w_{3}-3w_{4}$$. Then observe that $$\varphi\left(w\right)=\left\langle w,z\right\rangle$$ where $$z=\left(1,-3,3,-3\right)$$. The theorem I described earlier from linear algebra tells us that this $$z$$ is unique and is the cross product $$v_{1}\times v_{2}\times v_{3}$$. I hope this helps.

A good exercise is to use this definition for $$\Re^{3}$$ and try to prove all the things you know about the cross product from elementary calculus, like for example, that $$e_{1}\times e_{2}=e_{3}$$ or that $$u\times v=-v\times u$$. You can even discover why that symbolic determinant works for the formula for the cross product.

11. Dec 26, 2009

### AEM

First, Thank you for taking the time to come up with an example and the suggested exercise. Thanks also to D.H. for his post. One more question: How is this generalized cross product related to the wedge product of geometric algebra?

12. Dec 27, 2009

### cmendnba

Its really not that related. The reason some of the properties of the cross product are similar to that of the wedge product is that the cross product is defined in terms of determinants, and the determinant function is an alternating $$n$$-tensor on $$\Re^{n}$$.

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