Building invariants

Suppose we have a vector (contravariant) and we want to build an invariant.
a) we may take the direct product of the vector with some covariant vector (1-form obtained through metric tensor) and contract. The result is scalar.
b) we may take it's product with an axial vector (built with Levi-Civita symbol from antisymmetric tensor). The result is pseudoscalar.

I wonder if our vector may be acted on with some object giving result, that is neither scalar nor pseudoscalar, but having the following property:
with inversion of coordinates (in 3D) it acquires factor of e^{\imath \phi} ?
Does it imply necessarily that the metric should be complex?

Answers and Replies

Chris Hillman
Science Advisor
I guess you might be working toward the notion of a "covariant" as that term is used in invariant theory. See for example Peter J. Olver, Equivalence, Invariants, and Symmetry, 2nd ed., Springer, 1995.

Suppose we have a vector (contravariant) and we want to build an invariant.
a) we may take the direct product of the vector with some covariant vector (1-form obtained through metric tensor) and contract. The result is scalar.
b) we may take it's product with an axial vector (built with Levi-Civita symbol from antisymmetric tensor). The result is pseudoscalar.

I wonder if our vector may be acted on with some object giving result, that is neither scalar nor pseudoscalar, but having the following property:
with inversion of coordinates (in 3D) it acquires factor of e^{\imath \phi} ?
Does it imply necessarily that the metric should be complex?

If something acquires a factor $$e^{\imath \pi}$$ after inversion then after a double inversion, this something should not change, so it seems we must have $$\phi=\pi$$. So something just changes sign. So it seems that you are asking for a pseudoscalar (a quantity that changes sign under inversion of coordinates).

Last edited:
Thanks for your answers!
The particular case of \phi=\pi indeed gives pseudoscalar. But I've stated the question in the full generality. Double inversion ought not to be a "return to innocence" and some phase factor may be acquired (as in the case of two-dimensional spinors). The question is whether this behaviour can be expanded for any real numbers (and not necessary integer or rational multiples of \pi).
BTW what is the tag for LaTeX?
Thanks again

The LaTeX is entered as [ tex ] \ alpha [ / tex ] (remove spaces).

I think you are trying to invent a generalization of spinors. A spinor changes sign when you rotate by $$2\pi$$. But a spinor is not multiplied by $$\exp(\imath \pi/2)$$ under rotation by $$\pi$$. And also inversion just leaves a spinor invariant (as far as I remember) because inversion is not equvalent to a rotation (in three dimensions).

I don't think there can exist objects that are multiplied by some weird number under inversion. The reason is that you want a group of transformations, say O(3), to act on your objects. In other words, you want a representation of O(3). Representations of O(3) are known and classified, and there are no such beasts among them.

Now you understand me :-)
But, I am not sure, that no new irreducible representation can be proposed.
Everything depends on what kind of object is assumed to represent a physical state. If it is not a vector...
Let me think awhile and thanks for your remark
Thanks for the tag $$\imath$$

Hurkyl
Staff Emeritus
Science Advisor
Gold Member
Now you understand me :-)
But, I am not sure, that no new irreducible representation can be proposed.
Everything depends on what kind of object is assumed to represent a physical state. If it is not a vector...
Let me think awhile and thanks for your remark
Thanks for the tag $$\imath$$
If you don't want to get into category theory, then a group action on a set is the most general generalization of a group representation. (we call such a thing a G-set)

Let 1 be the identity element of your group G. Then, 1(x) = x, for any element x of any G-set.

If you have an element t of G satisfying tt = 1 (such as the element of O(3) that swaps the x and y axes of the canonical representation on R^3), then t(t(x)) = x, because t(t(x)) = (tt)(x) = 1(x) = x.

So, for any sort of group action of O(3), if you have an object and perform an inversion twice to it, you must get the object back.