Is B^{ij}_i a Contravariant Vector?

[cristina89]

Homework Statement

B is a third order tensor. Show that $B^{ij}_{i}$ is a contravariant vector.

The Attempt at a Solution

Well... I just thought about a simple solution but I don't think I'm right. But anyways.
Considering $B^{ij}_{i}$. If I raise the index i: $g^{ij}B^{ij}_{i} = B^{ijj}$
And so I can say that B is a first order tensor = vector. And this is a contravariant vector.
Is this right?

 

[HallsofIvy]

Homework Statement

B is a third order tensor. Show that $B^{ij}_{i}$ is a contravariant vector.

The Attempt at a Solution

Well... I just thought about a simple solution but I don't think I'm right. But anyways.
Considering $B^{ij}_{i}$. If I raise the index i: $g^{ij}B^{ij}_{i} = B^{ijj}$

You can't have the same index three times! Also, using the "summation convention" you sum over one "upper" and one "lower" index so even if you wrote $g^{kj}B^{ij}_k$, you would not be summing. Perhaps what you mean is $g^k_jB^{ij}_k= B^i$

And so I can say that B is a first order tensor = vector. And this is a contravariant vector.
Is this right?

Well, what is the definition of a contravariant vector? Can you show that this satisfies that definition? Remember that it is NOT enough just to show that something can be written with a single index- you can write an array, in a given coordinate system, indexed with a single index- that does not make it a "vector". A vector must change coordinates correctly as the coordinate system changes. If I remember correctly (it's been a while!) one test for a contravariant vector, written as $v^i$, is that the combination $g_{ij}v^iv^j$ (essentially the vector length) is a scalar. Which, again, does not just mean "a number" in a given coordinate system. A scalar (0 order tensor) must not change when you change coordinate systems.

 

[clamtrox]

Equally well you can show that $B^{ij}_i$ transforms as a vector, ie. if I do a coordinate transform $x^{i} \rightarrow x^{i'}$, B should transform as
$$B^{i' j'}_{i'} = \frac{\partial x^{j'}}{\partial x^{j}} B^{ij}_i$$