What is the tensor product dependent on the field in an algebra?

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?

• 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. Oct 14 '20 at 0:44
• @MichaelMorrow Thank you very much for the interesting pdf! 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.

• 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. Oct 14 '20 at 4:04
• @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. Oct 14 '20 at 4:21
• 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 $i$? Oct 14 '20 at 5:10
• @Giulio: correct, I'm saying that $1$ remains a basis of $\mathbb{C}$ over $\mathbb{C}$. Oct 14 '20 at 5:41
• If you want a really fun exercise you can try calculating $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}$, as a $\mathbb{C}$-algebra. Oct 14 '20 at 5:41