Tensor product of two Group Representation

Question

I have a question regarding the definition of tensor product of group representation. I found two definitions: One, is from Wikipedia. The other is from J. L. Loday's book Algebraic Operads. Please note that I don't know anything about representations of groups. Kindly, try to give an answer in simple terms, if possible. Let $G$ be a group and $\mathbb{K}$ be a field of characteristic zero.

Group Representation:

A representation of $G$, called as a left $G$-module, is a vector space $V$ together with a left action $\cdot : \mathbb{K}[G] \times V \to V$ which satisfies the following properties: \begin{align*} &\text{Distributivity:}\quad g \cdot (v + w) ~=~ g \cdot v + g \cdot w \quad \text{and} \quad (g + h) \cdot v ~=~ g \cdot v + h \cdot v \\ &\text{Scalar compatibility:} \quad k (g \cdot v) ~=~ (kg) \cdot v ~=~ g \cdot (kv) \\ &\text{Group compatibility:} \quad (gh) \cdot v ~=~ g \cdot (h \cdot v) \quad \text{and} \quad 1_{\mathbb{K}[G]} \cdot v ~=~ v. \end{align*} for all $v \in V,~ k \in \mathbb{K}$, and $g,h \in G$. Here $\mathbb{K}[G]$ denotes the group algebra of $G$ over $\mathbb{K}$.

Note: Similarly we have the notion of a right $G$-module. And since $\mathbb{K}[G]$, in general, is a non-commutative, unital, associative algebra, one can see $\mathbb{K}[G]$ as a non-commutative ring with unity.

Tensor Product of Group Representation (Wikipedia):

Let $M$ and $N$ be two left $G$-module. Then their tensor product, denoted by $M \otimes_G N$, is the tensor product of vector spaces $M \otimes N$ with left action $\cdot : \mathbb{K}[G] \times M \otimes N \to M \otimes N$ being defined as $$g \cdot (m \otimes n) := (g \cdot m) \otimes (g \cdot n)$$
for all $g \in G$, $m \in M$, and $n \in N$. Then extending it linearly to whole $\mathbb{K}[G]$ and $M \otimes N$ defines the left action.

The distributivity of left action follows from its construction.

Question 1: How do we show scalar compatibility?

Since I think $k(g \cdot (m \otimes n)) \neq (kg) \cdot (m \otimes n)$, as shown in the computation below:

$$k(g \cdot (m \otimes n)) = k (g \cdot m \otimes g \cdot n) \neq k(g \cdot m) \otimes k(g \cdot n) = (kg) \cdot (m \otimes n)$$

There is a similar question Tensor Product of Two Group Representations (Action Well-definedness), but the answer there doesn't discuss scalar compatibility.

Tensor Product of Group Representation (J. L. Loday):

Let $M$ be a right $G$-module and $N$ be a left $G$-module. Then $M$ can be seen as a right $\mathbb{K}[G]$ module and $N$ as a left $\mathbb{K}[G]$ module (considering $\mathbb{K}[G]$) as a ring). Then their tensor product over the ring $\mathbb{K}[G]$ is the abelian group $M \otimes_{\mathbb{K}[G]} N$.

Question 2: But how do we turn it into a $G$-module (left or right)?

Question 3: Do these definitions have any relation between them?

Edit 1:

Vector Space structure on the tensor product of group representation (J. L. Loday): Following the answer provided by Qiaochu Yuan I am trying to define a vector space structure on $M \otimes_{\mathbb{K}[G]} N$ in the following way: since $M$ and $N$ are vector spaces over $\mathbb{K}$, the abelian group \begin{equation*} M \otimes_G N := M \otimes_{\mathbb{K}[G]} N \Big/ \Big\langle \{km \otimes n - m \otimes kn: k \in \mathbb{K}, m \in M, n \in N \}\Big\rangle \end{equation*} turns into a vector space over $\mathbb{K}$, with scalar multiplication being defined as follows: $$k (m \otimes n) := km \otimes n ~=~ m \otimes kn.$$

Question 4: Is the construction fine?

Induced Representation (J. L. Loday): (Algebraic Operads, Appendix A.2, pg 458)

Loday gave the following definition of induced representation of a group $G$. Let $H$ be a subgroup of $G$ and $M$ be a right $H$-module. Then one can construct a representation of $G$ (, i.e., a left $G$-module) called the induced representation of $G$, which is defined as \begin{equation} \mathrm{Ind}^G_H(M) := M \otimes_H \mathbb{K}[G] \end{equation} where $\mathbb{K}[G]$ is viewed as a left $H$-module through multiplication in $\mathbb{K}[G]$.

Question 5: Following the above vector space construction. We have that $\mathrm{Ind}^G_H(M)$ is a vector space. But how do I make it a left $G$-module?

Edit 2:

Following Qiaochu Yuan's second comment, I am trying to give a correct description of the tensor product of two group representations.

Let $V$ be a right $G$-module and $W$ be a left $G$-module. The tensor product of $V$ and $W$ is a vector space \begin{equation*} V \otimes_G W := V \otimes W \Big/ \Big\langle \{m \cdot g \otimes n - m \otimes g \cdot n: g \in G, m \in M, n \in N\}\Big\rangle \end{equation*} over $\mathbb{K}$, where $V \otimes W$ is the vector space tensor product of $V$ and $W$. Thus, generators of $V \otimes_G W$ satisfies the following identities: \begin{align*} (v_1 + v_2) \otimes w = v_1 \otimes w + v_2 \otimes w, &\quad v \otimes (w_1 \otimes w_2) ~=~ v \otimes w_1 + v \otimes w_2, \\ k (v \otimes w) ~=~ kv \otimes w ~=~ v \otimes kw, &\quad (v \cdot g) \otimes w ~=~ v \otimes (g \cdot w), \end{align*} for all $v,v_1,v_2 \in V$; $w,w_1,w_2 \in W$; $k \in \mathbb{K}$; $g \in G$.

Furthermore, $V \otimes_G W$ can be viewed as a left or right $G$-module, with the following left action $ \cdot: \mathbb{K}[G] \otimes V \otimes_G W \to V \otimes_G W$ being defined as \begin{equation*} g \cdot (v \otimes w) := v \otimes g \cdot w \end{equation*} for all $g \in G$, $v \in V$, and $w \in W$. Then extending it to the whole of $\mathbb{K}[G]$ and $V \otimes_G W$ linearly we get the left action which makes $V \otimes_G W$ a left $G$ module. Similarly, the right action is defined by $(v \otimes w) \cdot g := v \cdot g \otimes w$.

Edit 3:

As Qiaochu Yuan pointed out in the comments that in general $V \otimes_G W$ is not a $G$-module. But if $V$ is a right $G$-module and $W$ is both a left and right $G$-module. Then $V \otimes_G W$ becomes a right $G$-module via the right action $\cdot: V \otimes_G W \otimes \mathbb{K}[G] \to V \otimes_G W$ defined as $$(v \otimes w) \cdot g := v \otimes w \cdot g$$

Answer 1

These are just two different operations. By default, "tensor product of representations of a group" refers to the first one. The second one is used to define group homology (edit: ah, and also induced representations, as you mention), and it is only a $K$-vector space and doesn't have an action of $G$ on it.

There is no issue with compatibility with scalars. The definition of the action of $G$ on the tensor product only applies to elements of $G$, it's not supposed to apply to elements of $K[G]$. The extension to elements of $K[G]$ is by linearity, so we have

$$k(g(m \otimes n)) = k(gm \otimes gn) = (kgm) \otimes gn = gm \otimes (kgn) = (kg)(m \otimes n).$$

Edit: The first equation of your edit is not correct but yes, the vector space structure on $M \otimes_{K[G]} N$ is just the obvious one, $k(m \otimes n) = km \otimes n = m \otimes kn$.

I have no idea why Loday is working with right $H$-modules. Normally representations refer to left $H$-modules, in which case induction should be defined via a tensor product on the left, as

$$\text{Ind}_H^G(M) = K[G] \otimes_{K[H]} M.$$

Then the left $G$-module structure comes from the left $G$-module structure on $K[G]$.

Generally speaking I think you should not be trying to learn representation theory of groups from Loday, since that isn't the focus of the book anyway.