Wolframalpha offered a one line definition to contravariant vector which is a bit confusing to me

Contravariant Vector:

The usual type of vector, which can be viewed as a contravariant tensor ("ket") of tensor rank 1. Contravariant vectors are dual to one-forms ("bras," a.k.a. covariant vectors).

Can someone elaborate on what it means for a vector to be "usual" versus "unusual"?

What does it mean that it can be viewed as "ket"? Does that mean contravariant vectors are or aren't ket states? i.e. can contravariant vectors be written as $|v \rangle$?

  • 1
    $\begingroup$ If one fixes a vector space $\Bbb V$, then "contravariant vectors" (kets) are just elements of $\Bbb V$ and "covariant vectors" (bras) are elements of $\Bbb V^*$. If $M$ is a smooth manifold, "contravariant vectors" at $p \in M$ are elements of $T_p M$ and "covariant vectors" at $p \in M$ are elements of $T^*_p M$. $\endgroup$ – Travis Willse Jan 7 '16 at 0:09

Consider a $\Bbb C$-vector space $V$ ($\Bbb C$ just because it's easy to visualize). Now consider the set of all linear functions $f:V\to \Bbb C$. That set is called the dual space to $V$ and is denoted $V^*$. You can easily confirm (by going through each of the axioms) that $V^*$ is also a vector space. In fact if $V$ is finite dimensional then $V^*$ has exactly the same dimension as a $\Bbb C$-vector space. Here addition and scalar multiplication work exactly as they do for any functions: let $v\in V$, $f,g\in V^*$, and $k\in \Bbb C$ then $$(f+g)(v) = f(v) + g(v) \\ (kf)(v) = k(f(v))$$

So because any vector space has an associated dual space, we could call some particular element of either space a "vector". But that would be ambiguous. So people have given names to vectors in each vector space. Vectors in $V$ are sometimes called "contravariant vectors", "kets", or even just regular "vectors" while vectors in $V^*$ are often called "dual vectors", "covectors", "covariant vectors", or "bras".

And yes, we physicists will usually write kets with the notation $\lvert v\rangle$ and bras with the notation $\langle v\rvert$. We don't normally explain why we're always allowed to associate some unique bra $\langle v\rvert$ with any given ket $\lvert v\rangle$ like this, we just take the Riesz Representation Theorem for granted (at least my professors always seem to).

Going a little further we could consider a differentiable manifold $M$. Attached to any point $p$ in $M$ is a "tangent space" and a "cotangent space". We consider the tangent space $T_pM$ to be the space of "contravariant" vectors or kets. These kets are the vectors you get by taking directional derivatives at the point $p\in M$ (actually we usually consider the directional derivative operators themselves to be the vectors).

enter image description here


Then the cotangent space is the space dual to it so we denote it $T^*_pM$ and call its elements "covariant" vectors or bras (or any of those other names). These bras are just the linear functions $\alpha: T_pM \to \Bbb C$ (assuming $\Bbb C$ is still the base field).

| cite | improve this answer | |
  • $\begingroup$ Excellent explaination! $\endgroup$ – Shamisen Expert Jan 7 '16 at 20:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.