A couple of months ago I posted a question about how to formally define systems of units, and someone posted this fascinating response, explaining that each base unit can be the basis of a single dimensional vector space and we can then use the tensor product to model multiplication of base units to form composite units. That part makes sense.

The part I don't entirely understand is that they say we can use the dual space to model division of base units. e.g. If $M$ is the vector space for meters and $S$ for seconds, then the composite unit of $\frac{m}{s}$ can be modeled as $M \otimes S^*$. How does that work exactly?

A dual space, $V^*$, is just the vector space of all linear functions that map $V$ to $F$, yes and division isn't a linear operation, so why should there be a member of the dual space corresponding to the reciprocal function?

On a side note, I did post this question as a comment on the original post and the person responded by saying

An element $α∈V^∗$ can be multiplied with an element of $v∈V$ by means of $α⋅v=α(v)$. Now since $V$ is one-dimensional, every $v∈V$ is of the form $v=λe$ for some chosen non-zero $e∈V$. Now, define $v^{−1}∈V^∗$ by $v^{−1}(μe)=μ⋅\frac{1}{λ}$, which is clearly linear. Then $v^{−1}⋅v=v^{−1}(λe)=\frac{λ}{λ}=1$, which justifies the notation $v^{−1}$. Also, clearly $v^{−1}$ is the unique element of $V^∗$ with this property.

I still don't quite understand though. I assume $\lambda$ is meant to be a scalar, in which case, it seems like all that's actually happening is we're defining a linear function that simply multiplies by the reciprocal of a scalar. In particular, I don't follow how $v^{−1}⋅v=v^{−1}(λe)=\frac{λ}{λ}=1$. It seems like just the $\lambda$s would cancel out and we'd be left with $e$, not $1$.

  • 1
    $\begingroup$ No, that’s the point. The basic linear map acts on $e$ and returns $1$. (By the basic linear map I mean the basis $\{e^*\}$ for $V^*$ dual to the basis $\{e\}$ for $V$.) $\endgroup$ May 27 at 23:21


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