Denoting vector spaces

Question

Let $X_i$ from $1 \leq i \leq n$ be vector spaces. Now I want to denote a space $X$, such that each $x \in X$ is given by $x = (x_1,...,x_n)$, and $x_i \in X_i$. Similarly, for each possible vectors, $y_i \in X_i$, for $1 \leq i \leq n$, $y = (y_1,...,y_n) \in X$. How do I denote $X$?

Can I denote $X = \bigoplus\limits_{1\leq i \leq n} X_i$?

Answer 1

The short answer is yes, the notation $X = \bigoplus_{1\leq i \leq n} X_i$ is standard notation.

For a longer answer, there are quite a few different ways to combine many vector spaces into a single vector space. The notation you are using is called the direct sum. There are two kinds of direct sums, internal and external direct sums. (I will borrow from Hoffman and Kunze for much of this - see there fore more details).

First, let's discuss internal direct sums. For this setup, we have multiple subspaces $W_i$ of a larger vectorspace $V$. We can in general define the sum of these subspaces $W_1 + \ldots + W_n$ to be the subspace of $V$ spanned by the elements of the $W_i$. The internal direct sum is a special case of this where our subspaces are independent, that is

$$\alpha_1 + \ldots + \alpha_n = 0, \alpha_i \in W_i \implies \alpha_i = 0 \forall i$$

So the notation $\bigoplus_{1 \leq i \leq n} W_i$ is precisely the same construction as the usual sum $W_1 + \ldots + W_n$ with the added extra information that these subspaces are independent.

The external direct sum tries to serve the same purpose but when our vectorspaces are not all given as subspaces of a larger vectorspace. To do so, we can construct a vectorspace that contains all of our summands as independent subspaces. This large vectorspace is the product $\prod_{1 \leq i \leq n} W_i$ which is given by the cartesian product and operations are componentwise. Then we can view $\bigoplus_{1 \leq i \leq n} W_i$ as the internal direct sum in this larger vectorspace.

In the case of finitely many vectorspaces that are finite dimensional, we have $\bigoplus_{1\leq i \leq n} W_i = \prod_{1 \leq i \leq n} W_i$. The distinction becomes interesting when you pass to an infinite case. For example, $\bigoplus_{i \geq 1} \mathbb{R}$ is not the same as $\prod_{i \geq 1} \mathbb{R}$ since the element given by a constant sequence of $1$'s, $(1,1,1\ldots)$, is in the latter but not the former.

There are other ways to combine vector spaces (e.g. tensor products) but these fundamentally serve a different purpose, so I won't go into them here.

Answer 2

The notation you write is ambiguous. To provide an unambiguous response to your question requires clarification. To see this, it must be the case that there is only one interpretation of what you write. If all you write is "$X = \bigoplus\limits_{1\leq i \leq n} X_i$", then the denoted object may fail to exist for certain choices of the objects $X_i$. To be unambiguous, your notation must have the same meaning for all choices of

• the vector spaces $X_i$ and
• all standard semantics of $\bigoplus$ (either internal direct sum or external direct sum) such that both
• every $x \in X$ has at least one representative $(x_1, \dots, x_n)$ for $x_i \in X_i$ and
• every $(y_1, \dots, y_n)$ for $y_i \in X_i$ represents at least one $y \in X$.

So, how can this be ambiguous? Let $W$ be the vector space $\mathbb{R}^3$ with basis $\{e_1, e_2, e_3\}$. Let $n=2$, let $X_1$ be the vector subspace of $W$ having the basis $\{e_1, e_2\}$, and let $X_2$ be the subspace of $W$ having basis $\{e_2, e_3\}$. This choice of $X_i$ meets your requirement that each $X_i$ be a vector space. Now take $\bigoplus$ to denote the internal direct sum. Note that the vector space sum $X_1 + X_2 = W = X$ since the span of these two subspaces contains an entire basis of $W$. However, the internal direct sum requires unique representations, which fails for these choices.

Any $((a_1, a_2),(b_1,b_2)) \in X_1 \times X_2$ represents $(a_1, a_2 + b_1, b_2) \in W$ so every $(y_1, y_2) \in X_1 \times X_2$ represents at least one $y \in W$.

Any $(w_1, w_2, w_3) \in W$ has infinitely many representations in $X_1 \times X_2$ : $((w_1, \alpha w_2),((1-\alpha)w_2, w_3))$ for any $\alpha \in \mathbb{R}$. (Simply put, this is where the problem is: in an internal sum, we can arrange for the $X_i$ to span, but fail to be independent. The independence requirement also appears in Xabu's answer.)

Although this meets all the requirements you wrote, the nonuniqueness of the representations of the objects in $X$ by objects in $X_1 \times X_2$ leaves "$X = \bigoplus\limits_{1\leq i \leq n} X_i$" undefined. This only denotes an internal direct sum if representations are unique.

There are three ways to fix the nonexistence of the direct sum that you wish to denote.

1. Include the prior constraint that the $X_i$ have trivial pairwise intersection. In an internal direct sum, this forces the $X_i$ to be independent subspaces so the representations are unique and the internal direct sum exists.
2. Alter your constraint to "each $x \in X$ is given uniquely by $x=(x_1,\dots,x_n)$, for some $x_i \in X_i$". This forces the $X_i$ to be independent, so the internal direct sum exists.
3. Explicitly assert that your use of $\bigoplus$ denotes an external direct sum. Then the $X_i$ are not interpreted as subspaces of an enclosing space, so are automatically independent.

Other resolutions:

• If you intend to use the notation $x_i + x_j$ for various elements of $X \cup \bigcup_{i=1}^n X_i$, use the vector space sum, "$X = \sum_{i=1}^n X_i$". The vector space sum places no requirement of uniqueness/independence, so exists even if the $X_i$ are not independent.
• If you are not going to add various vectors together, the direct product, "$X = \prod_{i=1}^n X_i$" assembles tuples of elements of the $X_i$ but does not impose an additive structure on the results. Note that the direct sum and the direct product differ if your $n$ is not finite. (The infinite direct sum only includes elements that are finite sums of elements of the $X_i$. Applying the forgetful functor from the direct sum to the direct product, the direct sum is sent to tuples with only finitely many nonzero entries. The infinite direct product places no constraint on the number of zero or nonzero entries.)