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Let $F$ be a field and $A$ an $F$-algebra. By definition we have a canonical embedding $\iota_A: F \hookrightarrow A$; the question (arising from a discussion in comments to What are the properties of this new characteristic of mathematical objects?) is, if or under what conditions, and how, can we well-define a "scalar part" of general elements of $A$; that is, find a "natural" section of this embedding, i.e. an $F$-linear map $s_A: A \rightarrow F$ with $s_A \circ \iota_A = id_F$. Motivating examples are the well-known "real parts" of complex numbers and quaternions viewed as $\mathbb R$-algebras.

Now of course there are many such sections by basic linear algebra: For any $F$-basis of $A$ which contains $1_F = 1_A$, we can just take the projection onto that vector. But that is far too arbitrary. E.g. $\{1, 1+i\}$ is an $\mathbb R$-basis of $\mathbb C$, but with this construction, the "real part" of $i$ would be $-1$. In a way, what we need is a "natural" vector space complement of $F$ in $A$ (defining our map by being its kernel)

I can think of two conditions which for me would qualify such a map, or collection of maps, as "natural":

  1. Being invariant under all $F$-automorphisms of $A$.

  2. For any embedding of $F$-algebras $A \subset B$, the restriction of $s_B$ to $A$ is $s_A$.

Of course, both for field extensions and matrix algebras (so by scalar extension for all simple algebras, and further for semisimple ones) a very "natural complement" of $F$ (assuming good characteristic) is the space of trace-$0$-elements, or in other words, our "natural" section should be (a rescaled version of) the trace. Indeed, the discussion in the comments under the linked question suggests the following:

Attempt: For finite-dimensional $A$, such $s_A$ can be found as follows: Choose an embedding $r_n: A \hookrightarrow M_n(F)$ and (assuming $char(F)$ does not divide $n$) set $s_A(a) = \frac{1}{n} tr(r_n(a))$.

Question 1: Is this attempt well-defined, i.e. independent of the chosen embedding?

Question 2: If yes, does this construction satisfy the "naturality conditions" 1 and/or 2? If no, do you have a better one which would?

Question 3: Conversely, does either or both of the "naturality conditions" force a certain construction on us?

(Bonus Question 4: What about the case of bad characteristic, and what about infinite dimensional $A$?)

Note: I am vaguely aware of tensorial and categorical definitions ("evaluation, coevaluation") of traces, but not very firm in those. If they turn out to be helpful for this question, I would be very happy about an explanation.

Torsten Schoeneberg
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  • In infinite-dimensional (and improper, such as a matrix containing infinities which case you seem not to mention) case, the trace may be infinite. Then depending on our purpose we can choose to regularize it, for instance, representing partial trace divided by partial rank as a sequence and regularizing it at infinity. – Anixx Jun 19 '22 at 08:45
  • For the trace method the matrix should be real. – Anixx Jun 19 '22 at 08:51
  • One example of an infinite-dimensional case is the space of functions $f:\mathbb R\to\mathbb R$.

    I think there are two natural definitions of scalar part here:

    1. The value of the function at zero: $f(0)$.

    2. The mean value of the function: $\overline{f(x)}=\lim_{h\to\infty}\frac1{2h}\int_{-h}^h f(x)dx=\frac{f(-\infty)+f(\infty)}2$

     – Anixx Jun 19 '22 at 11:27
  • The choice of the definition depends on whether we consider the function just as a set of values on its domain of definition or as a power series. – Anixx Jun 19 '22 at 11:58
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    Possibly, the matrix should be of minimal rank that allows such embedding, because $\mathbb R\times \mathbb C$ can be represented either as $\left( \begin{array}{ccc} a & 0 & 0 \ 0 & b & c \ 0 & -c & b \ \end{array} \right)$ or $\left( \begin{array}{cccc} a & 0 & 0 & 0 \ 0 & a & 0 & 0 \ 0 & 0 & b & c \ 0 & 0 & -c & b \ \end{array} \right)$. In the later case the trace obviously would be different. – Anixx Jun 19 '22 at 12:40

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