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The way that the covariant basis was described to me was that we could represent any vector $\mathbf a$ as either $\mathbf a=a_i\mathbf e^i$ or $\mathbf a = a^i\mathbf e_i$ (with the Einstein convention). But as I look at things on google and Wikipedia, it looks like $\mathbf e_i$ and $\mathbf e^i$ are two different types of objects -- namely a vector and a covector. And thus I don't understand how $a_i\mathbf e^i$ could equal $a^i\mathbf e_i$. Is this just a property of $\Bbb R^n$? Or is there something here I'm misunderstanding?

user233727
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  • take a look at: http://math.stackexchange.com/questions/1068862/covariant-and-contravariant-components-and-change-of-basis/1082980#1082980 – janmarqz Apr 24 '15 at 01:57

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Any finite-dimensional vector space has an isomorphism between vectors and covectors. In particular, after choosing a basis $e_1,\dots,e_n$ of $\mathbb{R}^n$, one can choose $e^1,\dots,e^n$ so that $e^ie_j$ is $1$ exactly when $i=j$ and $0$ otherwise. Concretely, if the $e_i$ are the standard column vectors ($n\times 1$ matrices), then the $e^i$ are their transposes ($1\times n$ matrices), where the product $e^ie_j$ is a $1\times 1$ matrix.

Kyle Miller
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    So they're never equal in the strict sense, but because there's an isomorphism between vectors and covectors, there's exactly one vector(/covector) that behaves exactly the same as any covector(/vector) in its respective vector space? – user233727 Apr 23 '15 at 03:45
  • Yes: the bases are the same size for both $\mathbb{R}^n$ and its dual space, so they are isomorphic. The thing about $n$-dimensional vector spaces is that they are all the same. Interestingly, finite-dimensional is important: an infinite-dimensional vector space has a much larger dual space. – Kyle Miller Apr 23 '15 at 03:51
  • @KyleMiller "Any finite-dimensional vector space has a canonical isomorphism between vectors and covectors." This isn't true. What isomorphism are you thinking of? "Taking the transpose" doesn't work since it's basis dependent. – Oscar Cunningham Apr 23 '15 at 09:40
  • "Any finite-dimensional vector space has a canonical isomorphism between vectors and covectors" No, far from it! "Canonical" has a meaning, and there is no canonical isomorphism between $V$ and $V^$. But a finite-dimensional vector space equipped with a basis* is canonically isomorphic to the dual vector space equipped with the dual basis, yes. – Najib Idrissi Apr 23 '15 at 09:42
  • Take a look at this: http://math.stackexchange.com/questions/622589/in-categorical-terms-why-is-there-no-canonical-isomorphism-from-a-finite-dimens – Najib Idrissi Apr 23 '15 at 09:49
  • @NajibIdrissi Yes, but that doesn't help the OP much because it's not a special property of the dual basis. A finite dimensional vector space equipped with a basis is canonically iso to the dual vector space equipped with any basis. – Oscar Cunningham Apr 23 '15 at 10:28
  • @NajibIdrissi I had "vector space with basis gives a canonical basis for the dual space" in mind. I removed the word "canonical." – Kyle Miller Apr 23 '15 at 17:32
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You're right vectors and covectors are different, and so $a_i\mathbf{e}^i$ can't be equal to $a^i\mathbf e_i$. I don't know what your teacher meant. Also, the other answer isn't right in saying that:

Any finite-dimensional vector space has a canonical isomorphism between vectors and covectors.

Covectors just aren't the same thing as vectors.

Oscar Cunningham
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