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I'm interested in understanding a comment by Chapter 0 of McCullaugh (1987). To keep the question self-contained, I'll provide some context below. Thank you.


Suppose I have vector space $V$ with $\{e_1,\ldots,e_n\}$. Let $v\in V$ be represented by $\sum_{i=1}^nv_ie_i$ where $v_i$'s are scalars. Let $K$ be some invertible $n\times n$ matrix and define new basis vectors $$ \bar{e}_i=\sum_{j=1}^nK_{ij} e_j\iff (e_1\;\cdots\;e_n)K'=(\bar{e}_1\;\cdots\;\bar{e}_n). $$ Then $v$ can also be written as $\sum_{i=1}^n\bar{v}_i\bar{e}_i$ where $$ \begin{pmatrix} \bar{v}_1 \\ \vdots \\ \bar{v}_n \end{pmatrix}=(K')^{-1}\begin{pmatrix} v_1 \\ \vdots \\ v_n \end{pmatrix}. $$ Comment of interest (first paragraph, page 14):

...(that) basis vectors are arbitrary up to linear transformation effectively means that all calculations are independent of the choice of the basis. In particular, all scalars are invariant i.e. independent of the choice of basis.

Question: what does invariant mean here? The new coordinates $\bar{v}_1,\ldots,\bar{v}_n$ above are not the same as the original coordinates $v_1,\ldots,v_n$ which are scalars.

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  • $\begingroup$ I assume it's simply supposed to mean that a scalar matrix is unchanged by a change of basis. The $v_i$ do not come from a scalar matrix, so they are not invariant. $\endgroup$ – Dustan Levenstein Jun 6 '16 at 15:39
  • $\begingroup$ @DustanLevenstein Are you saying that if $v_i$'s are all the same, then they stay unchanged? $\endgroup$ – yurnero Jun 6 '16 at 15:54
  • $\begingroup$ among other things, yes. $\endgroup$ – Dustan Levenstein Jun 6 '16 at 15:58
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The comment as it is in your post is taken out of context: in your post, it seems to be a part of some considerations about vectors - in which case one naturally asks how could scalars be particular cases of vectors (exactly your question). In reality, this comment in its original context is a part of some considerations about tensors, and in this case things are clear, because a scalar means a tensor of type $(0,0)$ or, if you prefer, an element of $V^0 \otimes (V^*)^0$, which is after all isomorphic to the base field.

To clarify this with an example, consider $\Bbb R^2$ with the basis $\{(1,0), (0,1)\}$ and the scalar $3 \in \Bbb R$. Now perform some arbitrary change of basis in $\Bbb R^2$ - it doesn't matter which one. Does $3$ change in any way? No, it doesn't even feel this change of basis - and this is the simplest type of invariance under a change of coordinates: being constant. (The reason why scalars -i.e. numbers- are invariant under such changes of basis being that they do not, in fact, have coordinates. Think for yourself: what are the $(x,y)$ coordinates of the number $3$?)

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  • $\begingroup$ +1 You're right that I left out some context. My understanding of the tensor approach is that it allows one to frame calculations in a vector space as calulations in the base field. What trips me up is the phrase "In particular", which suggests some logical implication in the exposition of which I don't see any. If you can clarify some more, I'll really appreciate it. If not, I think your answer is a good enough starting point so that once I read enough of McCullaugh, this concern will eventually make sense to me. $\endgroup$ – yurnero Jun 6 '16 at 17:20
  • $\begingroup$ @yurnero: I'd say that your starting point ("the tensor approach [...] allows one to frame calculations in a vector space as calulations in the base field") is mistaken, no wonder that it leads you astray. One possible understanding of a tensor of type $(m,n)$ is as a mapping $T : \Bbb R^n \to \Bbb R^m$ such that $T(u_1, \dots, u_i + v_i, \dots, u_n) = T(u_1, \dots, u_i, \dots, u_n) + T(u_1, \dots, v_i, \dots, u_n) \ \forall i$ and $T(u_1, \dots, \alpha u_i, \dots, u_n) = \alpha T(u_1, \dots, u_i, \dots, u_n) \ \forall i$... (continued) $\endgroup$ – Alex M. Jun 6 '16 at 17:27
  • $\begingroup$ @yurnero: (continued) ...The other approach understands tensors of type $(m,n)$ as the elements of a certain quotient space of $V^m \times (V^*)^n$ - but since I don't know whether you feel comfortable with quotient spaces I won't mention it. In any case, both approaches are "isomorphic", the former one being more concrete and easier to grasp for a beginner. Don't worry, tensors are a stumbling block for many students, and I believe that this is the main reason why many tend to avoid abstract differential geometry (where tensors are ubiquitous). $\endgroup$ – Alex M. Jun 6 '16 at 17:35
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Of course, writing "all calculations" is misleading. What one can say is the following: Given two vectors $x$, $y\in V$ the sum $z:=x+y$ is a well defined element of $V$ which does not depend on the chosen basis. In the same vein: You obtain the the coordinates of $z$ by adding the coordinates of $x$ and of $y$, whatever the chosen basis is. Similarly, the vector $w:=\alpha x$ is a well defined element of $V$ which does not depend on the chosen basis. Furthermore you obtain the the coordinates of $w$ by multiplying the coordinates of $x$ with this very $\alpha$, whatever the chosen basis is.

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  • $\begingroup$ Thanks for your answer. I just want to clarify that I didn't downvote. I'm still comprehending Alex M.'s answer. $\endgroup$ – yurnero Jun 6 '16 at 17:08

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