I don't understand the following notation:

$$V_F := V \otimes_k F$$

First of all, I know that that the product is a bilinear operation, i.e. $A \otimes A \to A$, between elements of the vector space $A$ in the algebra, but $F$ is a field, isn't it? $K$ is indeed a subfield of the bigger field $F$ with the operation restricted like in the classical example of $\mathbb R$ and $\mathbb C$. I've found a similar question and answer for vector spaces, and it explains that

$V_K$ is spanned by symbols of the form $a \otimes v$

but there it is noted that

these rules are not enough to combine every sum into an element of the form $a \otimes v$.

Therefore here, in the more complicated case of an algebra instead of a vector space, I'm even more confused...

Secondly, is there a way to reconcile the above algebraic definition with a geometric point of view (e.g. Lie algebra in differential geometry)? Where they say

The set of left-invariant vector fields $\mathbb g$ with the Lie bracket [ , ] : $g \times g \to g$ is called the Lie algebra of a Lie group $G$.

is there an equivalent definition in, let's say, noncommutative algebra?

  • 2
    $\begingroup$ Elements of the form $a\otimes v$ indeed belong to $V\otimes_k F$, but not every element of $V\otimes_k F$ is of the form $a\otimes v$. The elements $a\otimes v$ are referred to as simple tensors or elementary tensors, and every element of $V\otimes_k F$ is a sum of simple tensors. $\endgroup$ Oct 14 '20 at 0:44
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    $\begingroup$ I highly recommend reading kconrad.math.uconn.edu/blurbs/linmultialg/tensorprod.pdf $\endgroup$ Oct 14 '20 at 0:46
  • $\begingroup$ @MichaelMorrow Thank you very much for the interesting pdf! $\endgroup$
    – Giulio
    Oct 14 '20 at 3:51

If $V$ is a vector space over $k$, $V_F = V \otimes_k F$ is a vector space over $F$ called the extension of scalars of $V$ to $F$, with respect to a fixed choice of embedding $f : k \to F$. It can be understood explicitly as follows: if $v_1, \dots v_n$ is a basis of $V$ over $k$ ($n$ can be infinite here), then their image in the extension of scalars $v_1 \otimes 1, \dots v_n \otimes 1$ (often just written $v_1, \dots v_n$ again) remains a basis of $V_F$ over $F$. So for example

$$k^n \otimes_k F \cong F^n$$ $$M_n(k) \otimes_k F \cong M_n(F)$$ $$k[x_1, \dots x_n] \otimes_k F \cong F[x_1, \dots x_n]$$ $$\mathfrak{sl}_n(k) \otimes_k F \cong \mathfrak{sl}_n(F).$$

(So far these are all just isomorphisms of vector spaces.)

If $V$ has the structure of a $k$-algebra (commutative, associative, Lie, etc.) then $V \otimes_k F$ inherits this structure, but now as an $F$-algebra. The $k$-linear multiplication $m : V \otimes_k V \to V$ gets upgraded to an $F$-linear multiplication $m_F : V_F \otimes_F V_F \to V_F$. Again working explicitly, if $v_1, \dots v_n$ is a basis of $V$ and $m$ has structure constants

$$m(v_i) = \sum_{jk} m_i^{jk} v_j v_k$$

with respect to this basis, then the extension of scalars $m_F$ has structure constants $f(m_i^{jk})$ with respect to $v_1, \dots v_n$ regarded as a basis of $V_F$ over $F$ as above. This makes all of the isomorphisms I just wrote down above isomorphisms of $F$-algebras.

I don't understand your second question or what it has to do with your first question.

  • $\begingroup$ The problem I still have with this notation is that it depends on the choice of embedding $f$ which is not apparent in the notation: why? Thank you. $\endgroup$
    – Giulio
    Oct 14 '20 at 4:04
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    $\begingroup$ @Giulio: this is just for convenience. It's a standard "abuse of notation" to omit $f$. Usually it's clear from context, e.g. $f : \mathbb{R} \to \mathbb{C}$ the usual embedding for describing complexification. The construction also depends on other things like the choice of $k$-vector space structure on $V$ and we omit that too. We omit all sorts of things. $\endgroup$ Oct 14 '20 at 4:21
  • $\begingroup$ So $\mathbb R \otimes_R \mathbb C \cong \mathbb C $ and $V$ is $\mathbb R$ and $V_F$ is $\mathbb C$: how can a basis of $\mathbb R$ (I imagine it is $1$) remain a basis of $\mathbb C$ ? Is it still $1$ that now can be multiplied by any complex number? So is it different from a basis of $\mathbb C$ over $\mathbb R$ that would be $1$ and $i$? $\endgroup$
    – Giulio
    Oct 14 '20 at 5:10
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    $\begingroup$ @Giulio: correct, I'm saying that $1$ remains a basis of $\mathbb{C}$ over $\mathbb{C}$. $\endgroup$ Oct 14 '20 at 5:41
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    $\begingroup$ If you want a really fun exercise you can try calculating $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}$, as a $\mathbb{C}$-algebra. $\endgroup$ Oct 14 '20 at 5:41

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