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There are structures that are defined as various direct sums, like tensor algebras $T(V)$, exterior algebras $\Lambda(V)$ and so on. While all these structures are supposed to be external direct sums, which are cartesian products, they are almost always treated as internal ones, which are sums of subspaces.

And in each case summands are treated as subsets. While I can see how this works in case of internal direct sums, but it is clearly that for external case $C=A\oplus_E B$, neither $A$ nor $B$ are the subsets of $C$. It is more likely that $A\times\{0\}\subset C$ and $\{0\}\times B \subset C$, s.t. $(A\times\{0\})\oplus_I(\{0\}\times B)=C$

It confuses me alot, because it is often that operations defined on external sums, like wedge product, are used on summands though they have different underlying sets. Of course one can construct operation that will works on each summand from the given on the whole space, but that would be already different operation and we would need to distinguish between them.

Could you please clarify to me how this works?

Sergey Dylda
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  • Check out the answers to this question: https://math.stackexchange.com/questions/39895/the-direct-sum-oplus-versus-the-cartesian-product-times – Bernard W Apr 23 '17 at 04:42
  • @BernardWojcik I think it is not related to my question. – Sergey Dylda Apr 23 '17 at 04:49
  • Ok. Can you provide a definition of what you mean by internal direct sum and external direct sum? Or are you are asking for the definition? – Bernard W Apr 23 '17 at 05:00
  • For example in case of vector spaces: Given two arbitrary vector spaces $A$,$B$ one can construct a vector space name "external direct sum" as cartesian product $A\times B=A\oplus_E B$ equipped with pointwise operations: $(a_1,b_1)+(a_2,b_2)=(a_1+a_2,b_1+b_2)$ and similarly for scalar-multiplication.

    Then, given vector space $V$, and two vector subspaces $U$,$W$ one can decompose $V$ into "internal direct sum" $V=U\oplus_I W$ such that $\forall v\in V: \exists u,w: v=u+w$

     – Sergey Dylda Apr 23 '17 at 05:06
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    Mathematicians most often work up to isomorphism. It becomes second nature to view $A$ as a subobject of $A \oplus B$, because we are identifying $A$ with $A \oplus 0 \subset A \oplus B$ via the obvious isomorphism. – Alex Provost Apr 23 '17 at 05:12
  • @AlexProvost If we were to define some operation on $A\oplus 0$ then we would need to induce new operation on $A$ through isomorphism, aren't we? – Sergey Dylda Apr 23 '17 at 05:20

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I think that operations like direct sum are best understood first as internal operations. Let $R$ be some ring, with $M$ a left $R$-module. If I can write $M = N \oplus P$ as $R$-modules, this means that I have found a decomposition of $M$ such that $R$ will act on each component in an isolated way. In particular, any $m \in M$ may be written as $m = n + p$ for $n\in N$, $p \in P$, and then we have $rm = rn + rp$, where $rn \in N$ and $rp \in N$. Crucially, this composition does not "mix" under the action of $R$.

The external direct sum is then understood as a method of taking the modules $N$ and $P$ and constructing a module with the same behaviour of $M$ with respect to this decomposition, which symbolically just means replacing the $+$ sign in the above with a comma: $x = (n, p)$ and then $rx = (rn, rp)$. There is an obvious homomorphism from $N \oplus_E P \to M$, just be doing $(n, p) \mapsto n + p$, which is an isomorphism. So we may say that the module we cooked up is isomorphic to $M$, and so to some degree it doesn't matter whether we consider the direct sum $N \oplus P$ as being external, or sitting inside $M$.

Sometimes it may be easier to regard $M$ on its own, and other times it may be useful to extract factors out of the decomposition $M = N \oplus P$. If we want to go between these modes of thought a lot (and sweep the gritty details under the rug), we need an easy way of telling how an element decomposes into $N$ and $P$. For your examples, the tensor algebra and the exterior algebra have an obvious grading, being the rank of the tensor. If I write "The submodule of $T(V)$ generated by all $v \otimes v$", its obvious that this lives in the degree 2 part of the grading, which would be the $i = 2$ part of the exterior direct sum $$T(V) = \bigoplus_{i \geq 0} V^{\otimes i}$$ so most people will be happy going between internal and external direct sums based on the grading.

Joppy
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  • Thanks. When you wrote $(n,p) \mapsto n+p$ what's $+$ in it? On what space? Since we can't formally add element of different spaces. And if I want to speak formall of a tensor algebra, since it's the external sum, wouldn't it be fair to speak of "module that isomorphic to submodule" rather than just submodule? – Sergey Dylda Apr 23 '17 at 05:19
  • Here I consider $N$ and $P$ to be subsets of $M$, so addition is still defined in $M$. And no, it's better to speak of the exact submodule you want, because there could be many ways of decomposing a module as a direct sum. – Joppy Apr 23 '17 at 05:27
  • But how come $N$ and $P$ ended up being subsets if we constructed $M$ as external direct sum? I mean as in case of tensor algebra, we constructed bigger space from given ones via cartesian product but not the other way around. So, all summands are not the subsets of resulting algebra, but they are isomorhpic to the subsets of the algebra in which we can decompose our $T(V)$ – Sergey Dylda Apr 23 '17 at 05:34
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    $N$ and $P$ still have to come from somewhere (I need their underlying sets, addition rules, and $R$-action rules), and in this case I isolated them as being components of $M$ via an internal direct sum: $M = N \oplus P$. If I instead had $N'$ and $P'$ which were just isomorphic to $N$ and $P$ via the isomorphisms $f: N' \to N \subseteq M$ and $g: P' \to P \subseteq M$ respectively, then the external direct sum $N' \oplus P'$ would be isomorphic to $M$ via the isomorphism $(n', p') \mapsto f(n') + g(p')$. – Joppy Apr 23 '17 at 05:42
  • With the tensor algebra, there is no confusion between writing an element as $(0, v + w, v \otimes v, 0, 0, \cdots)$ and writing it as $v + w + v \otimes v$, because it's easy to see the grading, and go back and forth between both notations. – Joppy Apr 23 '17 at 05:46
  • I understand, I it is rather the question what is defined first. In case of tensor algebra we define summands first and then construct tensor algebra as external sum. So we can't just decompose it into subspaces right away, we still need to find isomorhism that will allow us to do so. – Sergey Dylda Apr 23 '17 at 05:51
  • However I'm more confused about wedge product on exterior algebra. Because it is defined as operation on the whole space, and if we want to use it on each summand we would need to induce new "wedge" using isomorphism $\phi$ as $\tilde{\wedge}=\phi \circ \wedge \circ \phi^{-1}$ – Sergey Dylda Apr 23 '17 at 05:55