Suppose $V^{r}_{s}$ is $(r,s)$ type tensor space over number field $F$. We can form an algebra by sum all these tensor spaces, namely $$T(V)=\oplus_{r,s\geqslant 0}V_{s}^{r}=F\oplus V\oplus V^{\ast}\oplus \cdots$$

My confusion is: How can we direct sum all these different vector spaces? As I know, direct sum is an operation among subspaces of a vector space. Are these $V^{r}_{s}$ subspaces of some large vector space?

  • $\begingroup$ You're describing an "internal" direct sum. This is an "external" direct sum. Mathematicians can be really lazy about making the distinction, and use identical notation for the two of them. While the external version is more general, they are essentially the same when both are possible. $\endgroup$ – JonathanZ supports MonicaC Nov 4 '19 at 14:20
  • $\begingroup$ @JonathanZ So the sum is just done formally? The meaning here is Cartesian product? $\endgroup$ – user450201 Nov 4 '19 at 23:49

Direct sum $V_1 \oplus V_2 \oplus ... \oplus V_n$ of finitely many vector spaces $V_i$ is the same as their cartesian product $V_1 \times V_2 ... \times V_n$. They are different, though, when one considers direct sums and products of infinitely many vector spaces.

  • $\begingroup$ Thanks for your answer. By the way, how does direct sum of infinitely many vector space look like? $\endgroup$ – user450201 Nov 4 '19 at 23:52
  • 1
    $\begingroup$ Cartesian product of $X_i, i \in I$ is the functions $ \{f: I \to \bigcup X_i | f(j) \in X_j \} $. Informally it works in the same way you can add and multiply by scalars componentwise. By the way, you can accept the answer. $\endgroup$ – Vladislav Nov 5 '19 at 6:26
  • $\begingroup$ A forgot to mention that direct sum of infinitely many vector spaces is a subspace in their cartesian product and consist of $a_i, i \in I$ such that $a_i \neq 0$ only for finitely many indexes. $\endgroup$ – Vladislav Nov 5 '19 at 8:58
  • $\begingroup$ For example consider $\mathbb{R}^{\mathbb{N}}$ (cartesian product of countable many copies of $\mathbb{R}$). It is the space of all sequences. But direct product of the same family is a subset in $\mathbb{R}^{\mathbb{N}}$ and consists of the elements of the form $(x_1, x_2, ..., x_n, 0, 0, 0, ...)$. $\endgroup$ – Vladislav Nov 5 '19 at 9:05

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